 okay so so we have changed the link actually so I have removed latency so now you can see clearly yeah so we are resuming it so just for a quick sorry for the glitch so it can just give me if you're there you are you know attendance if you're there so that I can start I can be a sure of sure of that you are there yeah so who all are watching okay sorry guys so there was this glitch now it should be fine so let's start our well let's resume so so we were dealing with the Thales theorem right so we dealt with Thales theorem I told you that basic proportionality theorem or Thales theorem is nothing but this that if there are two triangles sorry if there's a triangle ABC and a line which is drawn parallel to DE sorry a line D Pal is drawn parallel to BC then AD upon TV is equal to AD upon PC that's Thales theorem and the next one is the converse of this this is also called basic proportionality theorem so let's quickly revise that as well okay so the next one is if a line divides any two sides of a triangle in the same ratio then the line is parallel to the third side of the triangle so again if you see this is our triangle let's say we draw a triangle so if we have a triangle like that and we are saying that there's a if if if there's a line which divides this triangle ABC the sides of this triangle in such a way that that AD upon DB is equal to a upon EC EC then we say that we declare that DE is parallel to BC DE is parallel to BC this is converse of converse converse of BPD okay so keep that in mind now how did we prove that we used something called contradiction I'm not going again into deeper details contradiction will be used contradiction means let us say that given this ratio is there let us say that DE is not parallel to BC and we say that let another point DE dash be there on AC such that DE dash is parallel to BC then by con by our BPD we can say that AD upon DB is equal to AE dash upon E dash C why because this is just what we proved that if DE dash is parallel to BC then AD upon DB will be equal to AE dash upon E dash C but if you see the given is this this was given so that means from this relation and this relation we can say that AE dash by E dash C is equal to AE upon EC isn't it now that's if that's the case that means there is two there are two points on AC which divide AC in the same ratio that's not possible and hence we say that E dash must be equal to E that means E dash must coincide with E and hence and hence we say that DE will be parallel to BC that's how we proved converse of BPD okay enough now so let's go to the next one the next slide says the internal bisector this is called internal bisector theorem so these are the you know basic theorems you must remember the internal bisector of an angle of a triangle so let's say if there is a triangle like that ABC is this ABC is a triangle and there's internal bisector of an angle let's say angle A is being bisected such that this is D so A angle this is if this is angle theta this is angle theta let's say the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of contain of the ratio of the sides containing the angle that means if AD is AD is internal bisector internal bisector bisector of angle BAC if that is true then we know it is BD upon DC is equal to AB upon AC so this is these are the sides containing the angle AB and A so angle A is being bisected so hence AB upon AC is equal to BD upon DC so hence you remember like that AB upon AC is equal to DC yeah this is angle bisector of theorem again so how do you prove it so if you remember to prove this you have to extend AB you know in this direction such that AC is equal to let's say this point is E and then again using BPT if you if you really remember in this case we what we basically did was we extended we extended BA to AE such that AE is equal to AC now if AE is equal to AC and these two angles are same these two angles are same and and then if you or or what you can do is you can just yeah so if the so this side is equal to this side and this angle is equal to this angle right and if you see so it's given that AB by AC or yeah so basically we have to prove that AB by AE is equal to BD by DC then our job is done that means somehow you have to prove that AD is parallel to CE if this angle is equal or these these two sides are these two sides are equal or alternatively what you can do is you can produce CE in you can make CE parallel to AD and let let BA extended join CE at this point okay so using these so this is how you and then you using BPT you can prove that AB is equal to or AB by AC is equal to BD by DC so again since this is a revision class I will not go into the basics of it you can always you know revise through the text which we have already discussed in the class so I'll just skip these proofs so let's go to the next next proof okay or next theorem so next theorem is if a line through one vertex of a triangle divides the opposite side in the ratio of the two sides then the line bisects the angle at the vertex so basically it's converse of the previous theorem so if you see this is nothing but converse converse of the previous theorem converse of previous theorem what was that and what does it say so basically if again you have a line joining let's say ABC and then and then you have let's say this is BD okay so if AD is the point on BC such that it's given that BD by BD by DC is equal to AB by AC then then angle BAT will be equal to angle D AC this is what the theorem suggests so I would I would do one thing because we have lots of slides to cover so I'll what I'll do is I'll just try and recap the theorems first and then as time permits will come back and see how many proofs we can deal with okay now this is so you you know keep these theorems in one place as in we have already recommended that while solving problems keep these theorems in mind next is the line drawn from the midpoint of one side of a triangle is parallel to one of the side and if I six the third side it is a direct fallout of Thales theorem so let's let's again understand what exactly is this so there's a triangle there's a triangle and this is ABC and we say that line drawn parallel from the midpoint of one side so let's say this is E is the midpoint E is the midpoint of one side of a triangle and it's parallel to another side let's say we draw a line EF or let it be D instead of this being E let it be D okay so let is let us say this is point D and it's given that B is midpoint D is midpoint of AB okay D is midpoint of AB so what does it mean AD is equal to DB and then we are drawing a line parallel to line parallel to BC so clearly it you know you can understand from this direct result of Thales theorem if DE is parallel to BC then we know that AD by DB will be equal to AE by EC right by Thales theorem and hence the first part of this thing is proved so hence AD by DB anyways is one so hence AE by EC is equal to one and hence E is the midpoint as well E is the midpoint okay the next part of this is let me just clear the surface a little bit yeah next next next part of it is our staminate okay so and the next part I'll just remove this part but and yeah so the next part says that DE is now once E is midpoint so you say yeah the other part is DE is half of BC right and if you remember I'll just again I'll give you the hint why how it was done so you'd have expended DE to EF D to EF and such that DE is equal to EF then what will happen you will say that if DE is equal to EF AE is equal to EC anyways and this angle is equal to this angle so triangle AD E is congruent to triangle triangle CFE isn't it once that is done that means this angle is equal to this angle and hence this angle is equal to this angle so basically by this you will so basically DVC is equal to DF DFC these two angles are same and what else if you see so basically DF CB can be proven to be a parallelogram so this is a parallelogram how because this angle is is equal to this angle that means CF is CF is parallel to AD which is parallel to AB so CF is parallel to AB and AD is equal to DB is equal to CF why because by congruence we proved AD is equal to CF or FC and AD was anyway is equal to DB that means DB is equal to CF and we just proved that AB is parallel to CF that means DF CB is a parallelogram that means DF is equal to BC and DF was nothing but 2 DE is equal to BC so DE DE will be equal to half BC right so this is the next theorem next let's go to the midpoint theorem this is also called midpoint theorem okay now yeah oh sorry we missed this card the external bisector of an angle of a triangle divides the opposite sides externally in the ratio of the sides containing the angle so what does it mean this means simply this similar to internal bisector theorem we have external bisector theorem so let's say ABC is a triangle and BA is extended okay and let's say this is my so don't go on the you know scale it's not to scale actually yeah no worries so let's say this theta is equal to this theta and let's say this is a this is D okay so D is the AD AD is external is external bisector external bisector of angle CA and let's say this point is E CAE okay so if that is so then we know that AB upon AC AB upon AC is equal to is equal to BD upon BC okay that's what it will be right so how to remember this so so you start with AB and AC and hence you also start with from this point B so let's say AB is the first side AB and then B is the B is the starting point of the side on which the point B lies then you first stretch out from B to D and from D to C like that right so hence AB upon AC will be equal to BD upon DC like that okay so this is external and external bisector theorem okay let's move on so this is done cards a 8 7 is done yeah so internal angle bisector theorem external bisector theorem midpoint theorem now the line joining the two midpoints of the sides of a triangle is parallel to the third side again you know result of converse of converse of Thales theorem so you can prove this by converse of theories from ABC is a side and let D is the midpoint of D is the midpoint of D is the midpoint of AB and E is the midpoint of midpoint of AC then BE is parallel to BC BE is parallel to BC converse of this can be proved by converse of Thales theorem how so AD by DB so clearly AD by DB is equal to AE by EC is equal to one by one so hence since the ratios are same by converse of Thales theorem DE will be parallel to BC okay next so let's go on to yeah diagonals of a trapezium divide each other proportionally okay so let us understand this what is this diagonals of a trapezium so what is a trapezium trapezium is a quadrilateral having two sides parallel and the other two sides need not be equal okay if they're equal we call them isosceles trapezium abcd is a trapezium abcd is a trapezium and the diagonals of a trapezium so let's say diagonal AB and diagonal AC okay so this let's say this point is O so what does this theorem say the diagonals of a trapezium that means OD upon OB is equal to OC upon OA okay so they are dividing each other proportionally and again very easy to remember the proof what you do is you just draw a line parallel from O onto and join it with AD and let's this point be let this point be E okay so and then you can also yeah so from here you will get from here what will you get you will get in triangle in triangle DAB in triangle DAB what will happen by theories again by BPT by BPT since EO is parallel to AB then DE upon EA is equal to DO upon OB mind you guys I'm not writing all the statements because again this is a division session so we'll try to give you as many hints and you know you have already gone through it I'm just trying to recap everything so hence you should not be writing like these when there is a question on proof you should not adopt this methodology for all those questions previous yes questions we have we are doing a detailed model answers writing techniques so hence you can always visit those videos to see how the answer should be written this is just to recap whatever we have learned so that you can retain these during your exams and you can apply this so DE upon EA is equal to DO upon OB by Thelios theorem and by Thelios theorem again you in in triangle this was in triangle DAB similarly you can also take triangle yeah and then now you can try you can take a triangle ADC ADC in this also again AE upon ED is equal to AO upon OC AO upon OC so if you see these two using these two using these two you will see and you have to see for example here AE upon ED is there here it is the reciprocal of that ratio and hence if you reciprocate it it will become OB by DO and if you equate them you'll get AO this this particular relationship yeah so it falls out directly from here right so that's how you can prove this theorem I hope you guys are connecting to it if you want me to revisit any particular this thing you just let me know okay you can note down the card number if the time permits toward the end will always come back otherwise we can always take them offline and discuss it okay so let's go to the next slide next slide says if the diagonals of a quadrilateral divide each other proportionally then it is a trapezium again converse so let me yeah so this is the converse of the previous theorem itself so converse again this is the converse of the previous slide so what does it say it says that let's say there is a trapezium okay let's say there is a trapezium abcd and if they are let's say there's a diagonal there is a diagonal which is like that and yeah and this is O and is given that AO by OC AO by OC is equal to let's say BO by OD then then oh sorry this is a quadrilateral it's not a trapezium I'm sorry so there's a there is a quadrilateral and the diagonals are bisect sorry not bisecting dividing each other proportionally so this is how they will divide proportionally then then we have to prove that then we have to prove that abcd is a trapezium abcd is a trapezium so how do you do that okay so let us say that we are we draw a line parallel to so again you can take this as you yeah you take you find a point E such that if you find a point E on AD you can always find a point E on AD such that AE by ED is equal to AO by OC okay there's no problem so you can find a point E on AD such that AE upon ED is equal to AO upon OC and hence which is equal to BO upon OD because this was given anyways yeah so that means you have to take first this combination which combination so these two these two you have to take these two and then if you can if you take these two ratio then by converse of by converse of DPT you can say in triangle ADC in triangle so from this but follows in triangle ADC in triangle ADC what happens EO becomes parallel to DC isn't it by converse of Thales theorem and then if you take these two together that is these two together not these basically AE by ED is equal to BO upon OD if you take these two ratios together then what will happen you will see that OE is parallel to AB OE is yeah OE is parallel to AB and then you can combine these two to get what you can get AB parallel to DC that's what if you prove this that means that means ABCD the quadrilateral with two parallel opposite parallel sides that means it is a trapezium okay next so let us now go to next slide and this is okay so yeah any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionate now instead of diagonals now they are talking about let's say there is a trapezium and any line parallel to the parallel side so let's say if I draw a parallel line like that then they are saying if this is A this is B this is C and D then they are saying let's say this is D and sorry this is not D this is wait a minute yeah this is E and this is F and it's given that EF is parallel to AD is parallel to BC EF is parallel to AD one side and it's parallel to BC then they are saying that AE by EB is equal to DF by FC now how to do that so you can do it multiple ways and you know one is you produce the two sides of two non-parallel sides to meet at let's say O okay from here from here if you see you you get what again you know I said that I will not be covering proof but then quickly if it is possible let me just try and cover as many as possible so OA so in triangle OEF again OEF what can I write by BPT in this triangle OEF AD is a parallel line so I can say OA by EA is equal to OD by DF isn't it similarly in triangle OBC we can say OA by OB is equal to or rather OA by AB just a minute just a minute I'll delete yeah OA by AB OA by AB is equal to OD by DC okay now what you do is you divide one by two one by two if you divide one by two you will get AB by EA AB by AB by EA is equal to CD or DC whichever way DC by DF DC by DF isn't it now I am writing in this side and then subtract one from both sides you will get AB by EA minus one is equal to CD by DF minus one and if you if you solve it you'll get AB minus EA on AB minus EA on the left hand side which is nothing but so I'm just writing it here so it's nothing but AB minus EA so if you take the LCM and all you'll solve it you'll get AB minus EA upon EA now AB minus EA is BE so you'll get on the left hand side BE upon EA BE upon EA and on the right side CD minus DF is CF so you'll get on the right side CF upon CF on DF so if you see friends this is nothing but this or the reciprocal pattern right so hence you proved this as well okay so yes Ruby's asking some question I didn't get Ruby can you please type it there so that I can you know solve it in the second run okay let's move ahead okay if three or more parallel lines are intersected by two transversals then the intercepts made by them on the transversals are proportional okay I might be you know mistyping or mistyping and you know doing some this thing but you know you can highlight that and then you know we will do a correction so if three or more wait a minute again if three or more parallel lines are intersected by two transversals that means if let's say there are two or more parallel lines so they're saying there are oh sorry just a minute yeah so they're saying let me just delete this okay so there are three or more parallel lines let me draw some parallel lines and I will take this color yeah okay so parallel line one then another one parallel line two and then parallel line three like that okay and then it is intercept inter or intersected by two transversals okay so this is let's say this is one and this is another okay then the intercepts made by them on the transversal are proportional okay very good so what does it mean so let's say this is a b c d e f okay okay a b c d e f then they are saying a c upon c e is equal to b d upon d f yeah so this side upon this side or intercept is equal to this side upon this side which is again very easy to prove how you produce this you produce this and then again take triangle a c d first and then take triangle a e f with a b as the parallel parallel line a b parallel to c d so you first take this triangle which one a c d from that you will get let's say this point is o so from that you will get o a upon a c is equal to o b upon b d and then take triangle o c or o a e sorry o e f in that you will get o a upon a e is equal to o b upon b f and from whatever we just whatever we just proved in the previous card you can use the same to prove this i hope this should not be a big challenge for you guys okay same again so far we have been doing applications of bpt okay and just to add add here i will also tell you that in bpt there are multiple corollaries so hence you must be knowing and you must you can always quote this in your exams by proving sums so let's say a b c is the triangle and this is d is parallel to so d e is parallel to b c then all are true what all a d upon d b is equal to a e upon e c also a d upon a b is equal to a e upon a c also d b upon a b is equal to e c upon a c all these three are true you can do it by adding plus one minus one and things like that so there's a question that in card 13 okay let me go to card 13 so card 13 says wait a minute so in card 13 yeah so in card 13 can we just can we just construction and diagonal and prove it yes you can do that as well there is no problem so you can do that as well whichever way you want to do is that both are there would be multiple ways of proving the same question but question critica is asking is this let me share it with all other people so there are three let me just now delete this and in this card she's saying if there are three lines this is not parallel obviously it doesn't look like so this is let's say there are three lines parallel and there are two transversals like that so what she's saying is can I join these two and prove you can definitely prove how what is the proof so let's say all of you would be knowing it a cd and e f so again by application of vpt let's say this point is o so a c upon c e is equal to b o upon o e and in again in triangle b e f it will be b o upon o e is equal to bd upon tf both is this is also valid proof I hope I clear this next let's go on to our slides back yeah so so you could also do if you represent using the time yes yes param correct you're correct so you can use you can use thanks for adding to that okay so card 13 is done card 14 is done oh sorry I okay okay okay yes I was answering params question so critica you you're this thing also be done in the same way which I just proved for card 14 so the drawing which I did here was for part 14 so that's a correction next is okay now comes the second part what did we discuss we discussed these three parts isn't it so triangles basic proportionality theorem so far we learned around 14 14 theorems or let's say 14 piece of information around basic proportionality theorem it included midpoint theorem it included internal and external bisector's theorem so all should be there on your fingertips now we are entering the second realm and that similarity of triangles let's see how so first is the criteria for similarity so we learned criteria for similarity in some slide back and we said that corresponding angles must be equal and corresponding sides must be proportional isn't it so yeah so we for two criteria for similarity which has to be true it's another thing that we can we need not check everything for its validity but here here so we have some ways to reduce our checking criteria but then this will definitely be true always but all corresponding angles corresponding angles must be equal this is point number one and two is corresponding sides must be proportional sides must be must be proportional isn't it proportional so these are the two criteria now you don't need to check each and every criteria for you know checking the valid or let's say similarity of two triangles so hence we have some you know we you can eliminate your requirements some so not all the rest there so hence there is six criteria right there are three angles and three proportional sides okay so let me say so actually if you see you know that even if three all the three angles are same let's say this angle a is equal to d and b is b is equal to e and c is equal to f then also the two two triangles are similar and i will just try and do one proof for you arrest you can always you can use a similar method to prove all these criteria though the criteria usually they do not ask proving problems or criteria proving problems in in boards but you must know okay so hence if you see what do i do i will select a point e dash on a b such that a e dash is equal to a e dash is equal to t e so i select a point e dash such that a e dash is equal to d e and then i select another point f dash on on this thing um a c such that a f dash is equal to d f and then what do i do i join these so when i when you join this what will happen if you see triangle clearly triangle a e dash f dash is congruent to triangle d e f by s a s congruency criteria isn't it because angle a e dash this side is equal to this side and let me use another color so that becomes easier for you to understand okay okay and this side is equal to this side and boss this angle is equal to this angle so hence s a s criteria these two triangles are congruent so if these two triangles are congruent that means corresponding angles must be same that means e angle e is equal to angle e dash and angle f dash is angle is equal to angle f isn't it but my friend b is equal to e and e is equal to e dash that means b is equal to e so if b is equal to e dash that means line e dash f dash is parallel to bc right and if e dash f dash is parallel to bc then by using dpt that is basic proportionality theorem you can always say a e dash so hence the conclusion would be a e dash by e dash b is equal to a f dash by f dash c and hence you can and a e dash is nothing but d e so hence and and and the corollary to it will be nothing but a e dash by a b is equal to a f dash by ac and now you can replace a e dash by what d e and a f dash by d f so hence uh you know we we now say that even the two sides are two sides are proportional two sides are proportional now you can repeat the same process for angle b and then angle c and you will see that all such sides are all the corresponding sides are also proportional this is just one case there will be three such cases in all the three cases you can prove that the corresponding sides are proportional so this is the proof for let's say triple a similarity criteria so that means if in two triangles corresponding angles are equal then the triangles are similar fair enough so let's move on to our next slide okay so the next thing is a similarity criterion i don't need to go much into deeper details why because are because of and you know angle some property of a triangle they say that you don't actually need all the three angles to be checked if two angles are checked to be equal corresponding angles to be equal then you're done then the two triangles are similar a b c and this is d e f let's say d e f and angle a is equal to angle d and angle b is equal to angle e then automatically by angle some property what by angle some property which says that some of the three angles of a triangle is 180 degree so by angle some property angle c is equal to angle f and then by the previous slide now we have a similarity and hence the two triangles would be similar again so hence no worries this is clear to everyone i believe let's move on to the next slide okay triple a similarity criterion now they say that if if again so let me use this triangle only okay so let's say it says that triple a triple s is what if the three sides are sides are proportional that means a b a by d e is equal to b c by e f is equal to c a upon f d right then then then triangle a b c is similar to triangle d e f mind you guys while writing similarity please be very careful in writing the corresponding vertices against each other so triangle a b c need not be similar to triangle e d sorry uh f e d or just let me write it properly so need not be similar to um f e d so because corresponding angle a and f are not same a and d were same friends a and d must be written in the same same order okay so all all the vertices must be in the same order of equality that has to be taken care of okay so let's move on to next slide slide yeah now if one angle of a triangle is equal to one angle of another triangle and the sides including these sides are in the same ratio then the triangles are similar this is nothing but s a s this is nothing but wait and this is nothing but s a s so s a s s a s right so now wait right so yeah so wait so now um next is if two sides of sorry so let me go back with our right numbers like yeah if one angle of a triangle is this should be a triangle sorry so if one angle if one angle if one angle of a triangle is equal to one angle of another triangle and the and the side it's given that upon d e is equal to a c upon d f right if this is happening if this happens if this happens then we say that triangle a b c is similar to triangle d right this is nothing but s a s similarity okay so this is our another similarity criteria next now now now so just to summarize before we proceed further so how many similarity criteria did we learn one is one is a a a this is once then second is a a so these are these are mostly same you know third is s s s and fourth is s a s these are the four similarity criteria all right so please be mindful that one of whenever you are proving similar triangles one of these criteria has to be used okay fair enough let's move ahead and now see applications of similar triangles okay if you see you know application of similarity application of similar triangles and we'll see uh how these these applications turn out to or let's say produce these results so what is that first of all t i'll try and prove one and then rest you know doesn't take much of a effort to prove the next ones so if two triangles are equally angular that means uh they have same angles right if two triangles are equally angular then the ratio of the corresponding side is same as the ratio of corresponding medians so right so this is and others are so hence let me just first draw the triangles yeah so if okay before we proceed further i would just like to have a you know interaction with you so if you have any any doubt so far you please you can you know put it put forth here guys any any problem so far i'll just wait for your response as well any any problem so far guys please you can put it on you can you can just type in in the comment section so that i can i get to see or you can just you know simply say yes or no whichever way yeah so that i can see who all are there i hope you guys are not uh feeding board yeah so so later could you go back to card 12 uh okay angle by six to three boards without proving it another you know another problem okay uh i'll go to card number 12 once again no worries so let me see what was card number 12 but then you know i'm also running short of time but anyways okay let me just complete the entire thing i'll go back and then are we allowed to use angle bisector theorem in boards without proving it in another problem uh i would recommend not to do that if you're you know because i think you should not be using it because if at all you know or uh if if at all you have to require you know use it um then you should mention the theorem properly you know because uh you have in which lines don't just use it directly use you know you please mention the uh let's say the full uh the description of what angle bisector theorem is at least okay so i can see no other okay very good so let's move ahead so now we are on card number 19 and then it says if two triangles are equi angular then the ratio of this corresponding size is same as the ratio of corresponding mediums okay equi angular that means we have to take equi angular is nothing but equilateral triangle so let's say okay so hence let us take two equilateral triangle okay so anyways let's not attempt to make it equilateral so somehow this one and uh yeah something like that okay now the question is let's say if you have this as a b and c and this is d e f then what what are the same if two triangles are equi angular then the ratio of the corresponding side is the same as the ratio of corresponding medium so very okay okay so the ratio of the corresponding sides is same as the ratio of corresponding mediums so let me just draw a proper median here so this is the median so median what is the median median divides the opposite side you two equal parts okay so let's say this is g and h okay now it says that if two so hence a b it is given that a b by d e uh obviously if they're if the two triangles are similar or equilateral or equi angular then a b by d e is equal to b c by e f is equal to f b by ac is it now so just yeah so hence now if this is so then this is equal to a g by d h sorry yeah d this is g h okay okay and hence the other medians as well so let's say if this is i g h i and let's say this is j then also equal to me i by e j okay ac by f d okay so did i oh sorry i i wrote it the other way around thanks prajna for connecting so it is a b by d e a b by d e is equal to b c by e f is equal to c a i'm sorry i'm sorry this is wrong this is wrong and this has to be corrected thanks for correcting and this is c a upon f d okay now again um how do you prove so if these are equi angular then i don't you know uh this to this two angle is same this two angle is same and it's 60 degrees and b g is b g is nothing but half of b c and e h is equal to half of h f because a g and d h are median so hence if you see by angle side angle side again a b g triangle a b g triangle a b g let me write it down triangle a b g is similar to triangle d e h why because a b by e d is equal to half of b c by half of e f that is nothing but b g and b g upon e h and then angle a b g is equal to angle d e h isn't it so hence by what similarity criteria by s a s these two triangles are similar hence the corresponding sides will be equal to a g upon d h okay similarly corresponding sides is same as the ratio corresponding angle bisector segments what is angles bisector so in case of equi angle equi angular triangles angle bisectors altitudes also center in center or all sorry not centers altitudes medians your internal angle bisector and perpendicular bisectors all are same okay so hence this will be true for all of them this will be true for all of them you can try so what did we learn one is median what is median for your this thing median is nothing but line segment joining vertex to vertex to opposite side opposite side of the triangle this is opposite side midpoint opposite midpoint of already right midpoint midpoint of opposite side of the triangle this is median then altitude what is altitude altitude altitude altitude is nothing but perpendicular perpendicular from one vertex one vertex to the other to the other side opposite side this is my altitude then there is something called angle bisectors segment angle bisectors it's very clear so angle bisectors are nothing but line segment bisecting the angles of the triangle and then there is perpendicular bisectors perpendicular bisectors these are four optical you know elements which we study so median altitude angle bisectors and perpendicular bisectors so angle bisector is nothing but this is angle bisector let's say this is the triangle so this will be my angle bisector so this angle is this angle and this will be my perpendicular bisector right so this side is equal to this side and okay so you know this let's move on next so if one angle of a triangle is equal sorry if one angle of a triangle is equal to one angle of another triangle and the bisector of these angles divide the opposite sides in the same ratio so then the triangles are similar what does it mean let's see so if one angle of a triangle so let's say two angles are there one right a b c it says what a b c d e f two triangles are there what is given it's given that angle a is equal to angle d okay one angle of a triangle is equal to one angle of another triangle and the bisector of these equal angles so let's say so you have these bisectors of these angles so what what does it mean this angle is equal to this angle this angle is equal to this angle bisectors of these equal angles divide the opposite side in the same ratio so it's nothing but the converse of converse of our internal bisector theorem so if one angle of a triangle is equal to one angle of another triangle okay a is equal to d and the bisectors of these equal angles that means this you know a b c and then this is e d e f g let's say and this is h okay so it is saying that a a g is bisecting so a g is the angle bisector and it is dividing the opposite side in the same ratio that means it's given that b g by g c is equal to e h by h f then the two triangles are similar so if i'm not wrong it may not always work for perpendicular bisectors as they don't pass through the opposite verse oh in card number 19 you're talking about just let me see what was that okay there's a doubt so it says if you triangle is equal angle then the ratio of the corresponding side is same as the ratio of the corresponding altitudes yeah yeah that's true because they are yeah so in the case of equi angle time i was just mentioning about all the four elements because those four elements are usually so i was just trying to recap the four elements of the triangle which are usually you know utilized in problem solving so yes it will not hold true for the perpendicular bisectors as it is rightly pointed out that perpendicular bisectors will not pass through the vertices very true correct thanks for correcting so card 22 is this and we were explaining this to you yeah so hence it's given that angle a is equal to angle d and b g by g c is equal to e h by h f it is given then you have to prove that a or basically triangle a b c so hence if that is true these two criteria are fulfilled then triangle triangle a b c a b c is similar to triangle d e f right no so hence what what did we learn from internal bisector theorem if you remember b g by g c is equal to a b by ac internal bisector theorem says that in a triangle if there is an internal bisector of our diagonal let's say a g is the internal bisector then a b by ac will be equal to b g by g c that's what we learned in that internal bisector theorem so hence a b by ac so here a b by ac is equal to b g by g c will be equal to similarly here will be equal to d e by d f sorry d f sorry so this is d e by d f by d f now a b by a b by ac is equal to d e by d f and angle a is equal to angle d right already given one angle it's given from here angle a is equal to angle d so by s a s criteria side angle side criteria those two triangles are similar okay so this is also done next so if two sides and a medium bisecting one of these sides of triangle are respectively proportional to the two sides and the corresponding median of another triangle then the triangles are similar let me explain looks little complex while reading it so what does it say so hence you should do like this two sides okay so two sides of and a median bisecting one of these sides of a triangle so let's first make one triangle one triangle is like that so let's say a b and c let's say two sides so let's say let's pick up a b and ac these two sides and one of these and a median bisecting one of these sides so let me pick up this ac and let's say this point is d okay so triangle are respectively proportional to the two sides so we'll have to have another triangle let's say another triangle like that it's proportional to let's say p q r and s right so it is saying a b by a b by ac is equal to if two sides and a median i'm not sorry i'm sorry this is not a b by ac it's basically a b by a b by p q a b by p q is equal to ac by pr and is equal to b d by q s if this is so then triangle a b c is similar to triangle p q r this is my problem so hence again again how to read two sides and a median bisecting one of these sides so i pick two sides a b and b c and median bisecting one of these two sides one of these two sides is let's say ac pick d as the midpoint so b d becomes the median so two sides and the corresponding median is b d so a b and then it is they are proportional to the similar setup in another triangle so hence a b upon p q is equal to ac upon pr is equal to b d upon q s then the two triangles are similar this we have to prove okay fair enough so how to prove how to prove uh these two things now so clearly a b by p q is given we have to uh prove that these two triangles as when when will these two triangles be similar you have to prove a b by p q is equal to somehow a d by uh uh it is already given that a d a sorry ac it's already given that a b by p q a b by p q a b by p q is equal to a c by pr it's given and by uh ac by pr is given so hence it will also be equal to ad upon ps ad upon ps why because ad is nothing but half of ac and ps is nothing but half of a pr half of pr okay so when that is true that means if you consider these two triangles which which two uh yeah so in in triangle so let me just write here in triangle in triangle abd and triangle p q s okay what's given a b by p q is already given equal to or we just prove it proved it that it is equal to ad by ps and by given also criteria also it is equal to b d b d upon q s it's here so ab by p q is equal to b d by q s and we just prove that is equal to ad by ps also here so that means by s s s these two triangles are similar which two these two triangles are similar when these two triangles are similar my friend then this angle must be equal so hence if these two triangles are equal then what will happen ab by uh so these two these two triangles are equal and the corresponding sides are anyways given to be equal why so if you see ab by p q is equal to b d by pr i'm sorry b d by uh no you can't take this one ac by pr if you see ab by p q is equal to ac by pr ab by p q is equal to ac by pr and these two angles we just proved equal so hence the two triangles must be similar okay so this is slide number 21 next let's go to next slide if two sides and a median by setting the third side of a triangle are respectively proportional now in the previous case what did we learn in the previous case it was uh in this was slide number this was slide number 21 so in the previous case in the previous case there were two sides and a median bisecting median bisecting one of these sides but in the next slide this is a median bisecting the third third side so if you see the next slide the question is if two sides and a median so again if two sides these are two triangles triangle abc and let's say p q r triangle abc and triangle p q r what does it say it says that if two sides and a median bisecting the third side of a triangle so this is let me pick ab and uh p q so let's say these two sides are corresponding sides these two sides are corresponding sides so ab and p q and ac and pr and the third median of the third side yep me bisecting the third side so this is the median let's say abc let's say this is d and this is s okay they are proportional then again the two triangles are similar that what does it mean it says that if if ab upon p q is equal to ac if ab upon p q is equal to ac upon ac upon what pr pr is equal to ad upon ps then then these two triangles triangle abc is similar to triangle p q r right so then triangle abc is similar to triangle p q r how to prove this so okay this is also proportional so i'm just making a symbol to to identify that these three sides are proportional now again how would you prove it so if you see if you can somehow prove two sides of the median bisecting the third side of a triangle are respectively proportional to the two sides and the corresponding median of another triangle then the triangles are similar so that means so how what what has to be proven if you somehow prove that either the corresponding angles are same or the third side that is bc by q r is also same or it is equal somehow you could see that bc is bc by q r somehow if you prove this then you are done then you're done so let's see how what what can be done now medians are proportional okay fair enough that means okay so what to do in this case so basically just to give me a moment yeah so he is saying something you can use triple s right in card 21 i card 21 just a moment what was triple s in card 21 to the third size corresponding medium of other i didn't get this ruhi so i'll i'll come back to that never mind so you will come to this doubt let me just cover this doubt this this thing first so this question gives you ab by p q is equal to ac by pr and it is equal to ad by ps then you have to prove that the uh the two two uh this thing is if two sides and a median bisecting the third side triangle respectively proportional two sides okay then the triangles are singular so how to do it so if this is the third side then okay um this is the median so this is being bisected here so bd bd is equal to dc and q s is equal to s r okay so what to do so how do we do go about it how what was the proof ab by p q is given equal fair enough and ac by pr is also equal ac by pr screw ac by pr and this is ad by ps these three things are given okay so how do we prove this okay so um yeah okay so basically uh yeah what either you have to do by you can do a construction here so it is it doesn't seem to be so let's say yeah if i extend the median to make it equal and then joins what do i get i get okay uh yeah hence if you that you just have to that's fine so now this side okay and uh this side is this and this angle is this equal to this angle that means the angle abd and it's a cd uh cd r congruent that means ab and c s okay so what okay and then uh yes so let's say fix it then similar how does it help me so ad by p ab by p q is equal to ad by s that's fine okay okay okay um okay so primarily if i extend this i when okay these two are also congruent then i can some bd by bd by how i have to prove that this by this okay now wait a minute i will extend this okay i will extend that just give me a moment okay so this is these two triangles are also congruent that's fine and then ad by or we need to prove it by wait and this okay i'm okay kind of let's say we are stuck but never mind we somehow so ab if i if i extend ad if i extend ad to de okay okay so so these two sides are these two sides are just close the door so are you guys there hello hello you guys here guys you're there you just respond by saying yes in the segment i can understand that you're there hello you're there guys you're there i'm afraid i don't know whether you're there hello you're there guys are you able to hear me i see a contact hello guys i'm afraid hello the screen is so now okay just a minute it will stable just just a minute get stable maybe because of some internet issue just hold on just give me a give me two three minutes two minutes at least i'll just refresh it yeah is it fine now hello i can't see your responses oh okay so my screen was frozen so thanks yes yes yes yes now thanks thanks guys okay okay there's a lot of in a lot of yes sir yes sir good sorry i was my screen was totally frozen so there was some maybe fraction yes so anyone could you know attempt this make it full screen give me a moment please uh full screen okay so this has to be made full screen yeah so i in case you are you know whenever you are uh in in if there is some kind of a frozen screen you just let me know through whatsapp that's fine okay so i was trying to go through it okay now let me do one thing quickly guys i'll let's park it for some it's not frankly speaking it's not getting stuck here so let's i'm not uh you know i'll just let me just cover the slides first and then slide number 12 and 22 are left so let me just first in the interest of time let's first cover and then come back so we'll park 22 and 12 okay next is the ratio the areas of two similar triangles is equal to the ratio of the squares of any two corresponding slides both sides square of corresponding altitudes medians and angle bisector segments so quickly okay uh okay fair enough so anyways we'll come back to this side no worry so let's say the ratio of two angle to ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding side so you have already done this multiple number of times so and quickly cap this one and then this one again and then and um yeah so yeah the thing is areas of two similar two ratio of areas of two similar triangles is equal to the ratio of any two corresponding sides so let's say a b c a b c and d e f okay ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides okay so it's very you know so uh what is that area of triangle a b c by area of triangle d e f d e f is equal to nothing but a b square by d e square is equal to b c square by e f square is equal to c a square by f d square right again how to prove that again you have to drop a perpendicular from here right d e let's say b c it is p and this is q okay and then this is a perpendicular you drop okay and then you know that it's very easy triangle a b c quickly i'll recap it triangle a b c divided by triangle triangle d e f into b c into a p upon b c into a p upon half into base into height that is e f into d q right now half half get cancelled now somehow if i prove that uh a p by d q a p by d q is equal to let's say uh or or or if you prove that somehow a b upon d e a b upon d e is equal to a p upon d q then we are done and it is easily you know uh easily noticeable how because if you see this angle is equal to this angle and a p and and this is 90 degrees so i'm writing here for the lack of space here so i'm writing here so let's say triangle consider triangle a b p and triangle d e a b p and d e q okay so what a b okay a b uh yeah so a in in this in these two triangles angle b is equal to angle c angle e sorry and what angle a p b is equal to angle d q e is equal to 90 degrees a p b is equal to d q is equal to 90 degrees that means these two triangles which two these two triangles these two triangles are similar these two triangles are similar if these two triangles are similar then what can i say i can say a b by d e is equal to AP, AP by DQ. And if you use this in this, so this is what we had to prove and hence this can be written as BC. Okay, so it was it was AB by AB by D is equal to AP by DQ and since these two triangles were similar, so this could be also equal to BC upon EF because they were similar triangles. So hence AB by AB by DE is equal to AP by DQ which is equal to BC by EF. Why BC by EF? Because AB by DE was equal to BC by EF because these two triangles were similar. So hence from here you can say this is BC square by EF square. Okay, now clearly you can see instead of BC by EF, you can write AP by DQ here. Why? Because BC by E is equal to AP by DQ. So this is also equal to AP square by DQ square. This is also done. Squares of corresponding media. So in previous slides, you saw that in if two triangles are similar, then they are currently proportional in the previous slides. So this is also done. We also saw this in the previous slides. So angle bisectors are also there. So if for two similar triangles, angle bisectors are also similar. Sorry, also proportional. Okay, clear. So this is card number 23. So this is another, if you remember, we were talking about, sorry, yeah. So here is similarity of triangle applications is lots of, we saw the lots of applications and including the ratio of the areas. Okay, now let's go to card number 24. Yeah. If the areas of two similar triangles are equal, then the triangles are congruent. That is equal and similar triangles are congruent. Okay, so very, this also follows out from, you know, so they are saying if the areas of two similar triangles, so do two similar triangles are, there are two similar triangles, ABC, ABC and PQR and their area is same. So area is same. That means area of triangle ABC is equal to area of triangle PQR. That means, that means area of triangle ABC divided by PQR1, which is equal to nothing but AB square by, AB square by PQ square is equal to BC square by QA square is equal to CA square by PR square. And all the ratios are one. That means from all this, you can say AB is equal to PQ and BC is equal to QR and CA is equal to PR. And this side is equal to this side, this side is equal to this side, and this side is equal to this side. So these two triangles are congruent. Okay, bye. SSS criteria. Okay, next slide is this. If a perpendicular is drawn from the vertex of the right angle of an angle to the high part, use then the triangles on both sides of the perpendicular are similar to the whole triangle. And so another, this is this, we have asked you in multiple occasions also. So there's a right angle triangle first. We have dropped a perpendicular from here. So that says if a perpendicular is drawn from the vertex of the right angle of a right triangle. So this is the right triangle, right angle dropped a perpendicular. Let's say this is A, this is B, this is C, and this is D. ABD is dropped, BD is perpendicular dropped on AC. Perfect. Then the triangles on both sides of the perpendicular that that means that they are saying that triangle ADB is similar to triangle ABC. ABC and they, it's together is similar to, so ADB is similar to ABC and triangle ABC is similar to angle ABC is similar to C, D, B, angle BDC. Isn't it? So hence if you see angle ADB, ADB is equal, is similar to ABC. Very easy. Why? Because in these two triangles, if you see, angle A is common, angle A is same. So what are angles we are taking, we are taking, we are trying to be taking this, this one, this one. And let me choose another color to highlight it in different colors. So let's say I'm choosing this one, these two triangles. Okay, so I'm trying just the outline of it. So if you see this is common angle to both, this 90 degree is equal to this 90 degree, hence by AA similarity, they are similar. So angle A is equal to angle A, angle D which is 90 degrees is equal to angle B which is again 90 degrees. So hence both are similar. Now with, between ABC and CDB, same thing happens. So we are saying triangle ABC and triangle, and triangle CDB. These two. So hence common angle is C, common angle is C both. And D is perpendicular, angle D. This angle is 90 degree and this angle is 90 degrees. So hence B is equal to D. Both are perpendicular. So hence again this is similar. So hence they are similar to each other. If you remember, I had given you this question some time back in your module desk. Okay. Fair enough. So let's move ahead. We are running short of time. We have four slides more to go. So we'll complete. Okay Pythagoras theorem in a right triangle, the square of habit news, this you have, this is so many times you would have done. So let me not go again. So if you want, again, this is what the proof looks like. Pythagoras theorem says AC square is equal to AB square plus BC square. You have to use what? First of all, you have to use triangle ADB is similar to triangle ABC. And then using this, you have to find out one, one relationship between ADs, let's say between those two sides, that is the corresponding sides, and then using the other two sides, that is triangle CDB is similar to triangle CBA. Using these two triangles, these two similarity, you can prove that. I am not going in the details again. Yeah. So if triangle one is similar to triangle two and triangle three is also similar to triangle two, can we just say that triangle way and triangle three are similar or do we have to prove it separately? No, you don't need to. You don't need to. You don't need to prove it separately because it's clear, you're evident from, let me do it once again. So let's say I'm taking this. So triangle clearly, this is done when ADB is similar to ABC and CBD is, but you have to just maintain the order of the vertices. So let's say ADB is similar to triangle ABC is similar to triangle what? C remains at C's position, B goes to D and becomes BDC. Yeah. So, and this is how it is. Okay. So you can, you can do that directly. You don't need to prove them again. Okay. Next. So this is also done. So Pythagoras theorem, I am again leaving it on to you to do the next slide. Converse of Pythagoras. So Converse of Pythagoras theorem is again by contradiction, you can again use. So what is Converse of Pythagoras theorem? It says, it says that in a triangle ABC, if AC square is equal to AB square plus BC square, then AB angle ABC is right angle. I need to prove that. Okay. Yeah. So this is, okay. So this is also, how do you prove that? So let us say, you know, there are two possibilities either angle ABC. Let's say we start with contradiction again. So you can do that. Again, I'll just give it, I'm not going to do this again. So let's say if, let's say ABC is not equal to 90 degrees, then either ABC is greater than 90 degrees or angle ABC is less than 90 degrees. Let's say ABC is greater than 90 degrees. That means you can find a point C dash such that ABC dash, ABC dash is 90 degrees. Then you can write by Pythagoras theorem what? That AC dash square B square plus BC dash square. This is by Pythagoras theorem. And then by giving, you can write AC square is equal to AB square plus, plus what? BC square. Okay. And then you subtract both. You will get AC dash square minus AC square. Is equal to BC dash square minus BC square. Okay. And then from here, I'll use this. So from here, you can write AC plus into AC dash minus AC or AC dash plus AC is equal to BC dash plus BC into BC dash minus BC. Okay. Now, what do I learn? This is a positive quantity. So AC dash minus AC is negative. Okay. So this, this quantity is AC dash AC dash minus AC is clearly less than zero. Why? Because AC dash is smaller than AC. This quantity is negative. Now, this is also positive. That means that means BC dash minus BC must also be less than zero because this is positive. This is positive. This is positive. And if this is negative, then this is also negative. That means BC dash is less than BC. BC dash is less than, BC dash is less than BC. Okay. So I'm running short of time and space both. So hence, if BC dash is less than BC, that means, that means what? Opposite, so angle BC dash, so angle BCC dash is less than BCC dash, angle BCC dash, opposite angles should also be in the same ratio. So in this triangle BC dash C, you will BCC dash BCC dash, sorry BCC dash C. So basically this side is less than this side. So this angle will be less than this angle. Okay. This angle is less than this angle. So I'm writing here, don't mind. So that means angle BC dash C is less than angle BCC dash. Okay. Okay, BC dash C. So BC dash C is less than BCC dash. That means, that means what is this 180 degree minus angle AC dash, AC dash B is less than, is less than, yeah. So hence, 180, they see, there can be multiple, this thing, I'm just giving you one logical proof which can also be used. So 180 degrees minus AC dash B, this is my BC dash C is less than angle, less than angle, less than angle BCC dash. Okay. So, okay, yeah. No, so we actually wait a minute, wait a minute, wait a minute. So we actually, BCC dash is, BCC dash is, what did we do? BC dash is less than BC. So BCC dash is less than, oh, other way round, sorry, this, yes, this is other way round. Yeah. So this will be like that. So BC, BC dash C is greater than, I need some, you know, space here, wait a minute. So eventually, you will have to prove, eventually, what will have to be proven that this C, clearly, this angle is, BC, this angle is more than that. And yeah. So basically, what will happen? BC dash C is greater than angle BCC dash. So there is, yeah. So one, one, Mehul is suggesting you can even prove by constructing another triangle such that any two sides of the triangle is equal to any two sides. You can do that. Yeah. You can construct another triangle, but it will be similar to this approach, I think. Let us first try and resolve this because nothing should go unresolved. So hence I'm saying, if I think we are doing correctly BC dash, yeah. So AC dash minus AC is less than zero. And so hence BC dash must be less than BC, BC dash is less than BC. Okay. Now, yes, BC dash. So we have to prove that somehow BC dash is, BC dash is greater than BC. Somehow that, if that is proven, then we are done. Okay. So, okay. So let me just have some space too. So let me do it like that. Okay. Let me just, yeah. Don't worry, guys. I'll take just 10 minutes of yours and you're done. So basically, we are doing this. This is our triangle. There was small space. So let me just do it. Just bear with me for some time and we are done. So ABC, and let's say they are saying that AC square is equal to, AC square is equal to BC square plus AB square. Okay. And then they are saying that prove that angle ABC is equal to 90 degrees. Okay. Now we are saying, let's say it's not 90 degrees and angle, now we are saying, let us say, and ABC dash is equal to 90 degrees. Let us say, let's say, okay, then by this, what do I get? I get by Pythagoras theorem, AC dash square is equal to AB square plus BC dash square. And here it is, and here it is, given as AC square is equal to AB square plus BC square. So let me subtract both of them. So I'll get AC square minus AC dash square is equal to BC square minus BC. So that I can write as AC minus AC dash, AC minus AC dash into AC AC dash is equal to BC much into BC plus BC dash. Okay. Now clearly, this is greater than zero. This is also greater than zero. AC minus AC dash, that is CC dash, this is also greater than zero. That means, that means BC minus BC dash is greater than zero. Okay. BC, BC minus BC AC minus AC dash is greater than zero. And this is positive. So BC square. Yeah. So BC minus BC dash is greater than zero. So I have to just contradict this. If BC is greater than, if BC is less than BC dash, my assumption was wrong. This assumption was wrong. So let's try and prove that. Now clearly BC, BC minus BC dash is greater than zero. That means what angle or BC is greater than BC dash. This is what that means. That means, that means, what does it mean? It means angle opposite angle. So opposite angle is BC dash C is greater than, is greater than, is greater than angle BC dash opposite is BC C dash. Right. So BC dash C, BC dash C is greater than angle BC C dash. Okay. Now BC dash C can be written as 180 degrees minus 180 degrees minus what? AC dash B. And this is greater than angle BC C dash. Okay. Now, so that means, that means what? 180 degrees is greater than angle AC dash B plus angle BC C dash. Now check what is AC dash. If you see angle AC dash B, angle AC dash B is this angle. I can't show so the angle is greater than AC dash B plus BC C dash. So this plus this, isn't it? So this is what they are, see how, how, because yeah, so 180 degree angle is greater than 180 degree angle is greater than, sorry, guys, it's getting a little delayed. So this is equal to angle AC dash B can be written as angle two. I think we are heading the wrong direction. So just let me see what all are left and then, yeah, sorry for interruption once again. So we'll just cover inverse of Pythagoras theorem is there. So what all slides are left? I think three more theorems. Yeah. So these three theorems probably we'll take up in the next class, but we cover cover at least the, the theorem part of it. Okay. So that angle got little deviated. Let me just see. I'll just try and prove using our conventional method. So converse of converse of this thing is very simple. It didn't work. So let me just try this. So what the Mehul was suggesting is you have to draw another that's a conventional method. So let's see, since it's a board example, let's not try. So let's say A, B, A, B, A, B. A, B, C is the, A, B, C is the, is the triangle where A, B, AC square is equal to AB square plus BC square. Okay. And then we have to do what to prove that angle B is equal to, angle B is equal to 90 degrees. So what you do is you draw another triangle. So this is your NCR to prove maybe this D, E, F. This is E, F you draw such that what is that? D is equal to AB. So AB is equal to DE and EF, EF is equal to BC or BC is equal to EF and angle E is equal to 90 degrees. So you make it like that. You make our triangle equal to 90 degrees. Now, clearly if in this triangle, this is construction, construction. So if you do that, you will get what D F square is equal to DE square plus EF square, right. And D F square hence can be written as AB square plus BC square and BC function F. So hence, when I'm using this, you can write, but AB square plus BC square was AC square. So D F. So AB square plus BC square it's given from here is AC square. So write AC square. So hence, D F is equal to AC. So hence, in both these triangles, in both these triangles, if you see AB is equal to DE and BC is equal to EF and AC is equal to D F. Hence, angle E is triangle E E F. Hence, hence, the C is equal to which is E E F. This is a conventional proof. So if at all there is this proof you can use this method. So hence, construct another triangle with corresponding these equals, these equals and then AC square is equal to, you know, you can do that. Prove the two triangles to be congruent and then prove that this is equal to 90 degrees. This is how the conventional proof goes. I will again, you know, we in the class, we attempted the another proving method will again solve and send it back to you. I'll just read out these theorems so that your this thing is complete. So hence, in any triangle, the sum equal to twice the square of half. This is we have done in the class as well. So it says if this is a triangle ABC, this is actually called a Polynas theorem. So in any triangles of the squares of any two sides, so AB square plus AC square is equal to the twice the square of half the third side and the median, which so hence it is equal to twice the square of twice the square of half the third side. So this is D. So two times DC square plus AD square. This is a Polynas theorem. We have discussed the proof in our classes, right? This is a Polynas theorem. Again, you have to use Pythagoras theorem to so what you need to do is you need to drop up a perpendicular here. Drop up a perpendicular ABC, let's say put it P and then use Pythagoras theorem and express AB square in terms of AP and BP express AC square in terms of AP and PC and then and then express. So there are three Pythagoras theorems. You'll write you'll write AC square equations. You'll write AC square is equal to AP square plus PC square and then you'll also write AD square is equal to AP square plus PD square and then you'll write AB square is equal to AP square plus BP square and using these three you eliminate AP and you know another equation is B is equal to BC is equal to half of BC. Using all this you eliminate AP and every point related to P and you will get this. So this was a Polynas theorem proof. Now next we are almost done. So let's go to card number 29. Three times the sum of the square of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangles. Yes, this is again a very important theorem. You need to know this but then you avoid using them directly in your problem solving. Actually they will there will not be any direct application but yeah. So what is this theorem? This theorem says that okay this is ABC and it says three times the sum of the squares three times AB square plus BC square CA square is equal to let's say these are the medians and this is the direct application of the previous theorem that is the Polynas theorem. So let's say this is DEF is equal to four times BD square BD square plus AF square plus EC square okay AF square three times AB square plus BC square plus CA square is equal to four times BD square plus AF square plus EC square. How to again use this Polynas theorem? Polynas theorem says what AB square plus AC square is equal to twice of AF square plus AF square plus BF square and similarly you times and then eliminate so AB square and then you write BC square plus CA square is equal to two times BC CA is two times CE square plus BE square and then again you write third one this is one this is two and then third word I am writing here and that is nothing but CA square plus AB square CA square plus sorry not AC plus AC we wrote and BC and CA I wrote then I have to write AB square plus BC square sorry this is BC square so BC square plus AB square BC square plus AB square is nothing but twice of BD square plus BC square using these three using three add all of them together you will get this relation this relation by saying that and you know that how do you replace DC so DC is nothing but CD is nothing but half of AC you write that and BF is nothing but or BF is equal to wherever BF is there is BF so BF can be written as half of BC and and BE is equal to half AB so using these three and these three also together you will get this relationship you know just eliminate fair enough so this is card number 29 so this is also done and what next I left with this one three times the square of any side of an equilateral triangle is equal to four times the square of the altitude this is again fallout of the previous one so hence if you see this is nothing but if there is an equilateral triangle equilateral triangle so there is an equilateral triangle this is equal to triangle and they are saying that three times the square of any side of an equilateral triangle so let's say ABC right so three times AB square is equal to four times the square of the altitude four times the sign for AB city and you can see his direct direct trip per person or direct fallout of the previous one why the previous one was this three times what AB square three times the square of any side of an equilateral triangle yeah so AB square plus BC square plus AB square plus BC square plus what CA square is equal to four times the median and in this equilateral triangle case median is equal to altitude so four AD square plus let's say AD BE this is BE BE square plus CF CF square right now in in case of equilateral triangle AB is equal to BC is equal to CA right all sides are equal and altitudes and medians are also equal so you can write AD is equal to BE is equal to CF using these three what can you say from here so let's like this color yes so if from here okay if from here what will happen it is nothing but three square is equal to four times AD square will be equal to BE square equals to CF square from here so hence three three gets cancelled so hence three AB square equals four AD square right so this is your summary of the entire this you know now what remains is problem solving obviously these are the you know points to you know keep in your mind while you are you are trying to solve problems so beyond these no theorem and no you know application will be there all the questions you will be getting you will be having these only theorems you know useful now always keep in mind I told you what did I tell you that just a minute I'll show you yes so this is the basic information this one yeah basic information says that the moment you sequence a little triangle what could come to your mind first use of ways of similarity or in all probability one or max a couple of you know things to prove related to this thing theorem to be related to stuff otherwise mostly that if there will be two there will not be a direct theorem there will be indication of these so always keep in mind moment you see a triangle from e application of basic proportionality theorem for a similarity of triangles all the additional there no corresponding squares of sides and all that will be included so once you have the list of all theorem in front of you then you know you will be able to think through the problem and solve it so again there will be if the problem related is related to ratios or if you see lots of parallel lines in the diagram or if in the form of square so you the moment it is in the form of square then you know either the application of Pythagoras theorem or it will be the ratio of areas ratio of areas these are the only two techniques through which this can be solved okay guys so if you have any queries queries questions you can always reach out to me and that we are we are you know recording a lot of problem solving videos which are in progress if you want any particular problem to be solved and send back to you you can always you know reach out to me or any of any of the other faculty members or you can also you know mention in the comment section also no problems session usually you know I actually got a little bit prolonged no worries we'll try to kind of do our session from but I hope it was useful so next class I believe is on Friday and has been circulated to all of you please try to attend the session so that it becomes a big division of all that is there at least it will give you an overview of the entire process which is there and it will be there in your there in your mind okay so yeah so let's call it today and then we'll meet again thanks for your time and keep watching the problem solving videos and keep solving worksheets we have shared a lot of content material in the drive as well so please utilize them try to solve as possible this is the only way you can improve your score now okay thanks a lot so we will close here thanks for your time good night