 is my voice audible ok just write in the comment box what are your doubts which chapter I should start with which chapter I should start with guys I'm just asking you tell me your preference ok quadrilaterals I'll give you few questions of quadrilaterals try solving it so the question is show that quadrilateral formed by joining midpoints of the consecutive sides of the rectangle is rhombus solve it so in this question try to find out properties of rhombus you all two key words are there one is rectangle second one is rhombus and the third important point is joining midpoints so first draw rectangle and then join the midpoints and name them a b c d e q r s now what is rhombus all four sides should be equal that's how we have to do it just try to find out I've already made that diagram ok dhruti has done ok done tana has done shraddha are you done ok so just for the sake of confirmation what you need to do is you need to draw this construction needs to be done so I am drawing bd bd is diagonal of rectangle a b c d now if I take triangle d c b so what happens in this triangle that r and q are midpoint of dc and bc so r q would be parallel to db and r q would be equal to half of db similarly if you take other part of triangle which is d ab similarly you can write that ps would be parallel to db and ps would be half of db if I draw other side of diagram so from here you can say that opposite sides are parallel and equal so you can say that p q r s is a parallelogram now if p q r s is a parallelogram what I need to prove is I need to prove this that all sides are equal so what I will do is I know that opposite sides are equal in rectangle so I can write that ad is equal to bc so half of ad is equal to half of bc and similarly so what you can do is you can take you can take this triangle so you can take triangle s dr or I have already proved this and this equal so I can take this triangle s dr and I can take triangle asp so here half of ad is as and half of or I will take ad here and dc here instead of this thing so just give me a moment or you can take other side of the triangle also so what you can do is that this side if I take triangle ad r so I know that ad is equal to sd I also know that as ab is equal to dc so half of it would be ap is equal to dr and I know that angle a is equal to angle d so it would be sas congruency so sas congruency means side opposite to equal angles I mean sr would be equal to ps so sr would be equal to ps so if sr is equal to ps I have already proved that r q r is equal to sp now I am proving that sr is equal to ps it means that all sides are equal so you can say that this parallelogram is a rhombus write down the next question the next question is the diagonals of a quadrilateral abcd is abcdr perpendicular show that quadrilateral formed by joining midpoints of its sides is a rectangle done have you all done it okay two people have done druthi and two more minutes okay fine so most of you have done so I am giving you next question from quadrilaterals I don't know whether I have given this question before or not but still I want to solve it for revision purpose prove that line segment joining midpoints of the diagonals of a trapezium is parallel to each of the parallel sides is equal to half of difference between parallel sides do it quickly just make the diagram properly in this case if you make the diagram properly if the figure is in front of you you will be able to solve it make the diagonals and then join the midpoints of the diagonals so this is a this is b this is c this is d and midpoint suppose I call it b and then q and suppose this point is 4 now you will have to do some construction okay so let me solve this question I don't know how many of you have solved it okay druthi has got first part solve the second part also I am just waiting for that okay I will solve the second part so let me solve the first part only so first try to understand a b is parallel to dc so this angle suppose this angle is 1 and this angle is 2 so angle 1 would be equal to angle 2 because there would be alternate interior angles now I am drawing I am doing a construction so I am drawing a line dr through p so one thing is clear as p is the midpoint so let me write down what I know I know ap is equal to cp I also know bq is equal to dq now look at here if I take these what I have to prove I have to prove that this line is parallel to dc and a b so I am writing here to prove is pq is parallel to a b and dc and pq is half of a b minus dc these are the two things that I need to prove now how do you prove that one line is parallel to second line that's very important so try to understand if I am able to prove what what is the purpose of drawing this line why I am drawing this line through p the purpose of drawing this line to p is that I want to take this triangle drb is it okay and in this triangle drb if I am able to show that p is midpoint of you can take this from here also instead of dr you can make a point ct or something like that so what my objective is I will use the theorem of the triangle which theorem of the triangle that if I join midpoints of the adjacent side the line joining that midpoint would be parallel to the third side for that what I need to do I already know that q is midpoint of db but I don't know whether p is midpoint of dr or not so in this particular case I am writing again p here because it's not looking so what I need to prove is that I need to prove that this p is dividing dr into two equal halves you should not get confused that p is midpoint of ac hence you can simply write that p would be midpoint of dr also you will have to prove it here now how do you prove it here so to prove it here I should take two triangles which are the two triangles so let me take you should obviously know that it can be only proved that if I am able to prove that dp is equal to pr p would automatically become midpoint of dr so how will you prove dp is equal to pr for that you have to take triangle apr and you have to take triangle dpa this triangle and this triangle I am taking so how will I prove here I know that or you can do one more thing here it is not possible to prove these two lines you can do one more thing you can prove on from the other side you can take this triangle and this triangle you can take dpc so you can take triangle apr and you can take triangle dpc what was the problem in initial one that you only this line was equal no other angle was looking equal here the angles are looking equal so I can write angle one is equal to angle two these two angles are opposite angle so I am giving it a name of angle three and four so I am giving it name angle three is equal to four so these are alternate interior angles these are opposite angles and I know that AP is equal to CP so what kind of congruency it is it is ASA congruency so by CP CT you can write that angle approach sorry side opposite to angle one is PR and side opposite to angle two is a dp so PR is equal to dp if PR is equal to dp it means that P is midpoint of this now by midpoint theorem you can write that RB is parallel to PQ and RB is a part of AB so PQ would be parallel to AB AB is already parallel to DC so first part can be proved using now second part how do you do second part so just write it down then I will do second part have you written it okay I hope you have understood this thing how to do second part I will just explain you for second part what I need to do is that so try to understand P and Q are midpoints of DR and DB so PQ would be equal to half of RB now what is RB so this has been RB is equal to RB can be written as 1 by 2 RB can be written as AB minus AR and I know that AR is so what I proved that I just proved that triangle APR is congruent to triangle PDC I just proved it so look at here opposite to angle 3 AR is there opposite to angle 4 DC is there so I can replace this AR with DC so PQ is equal to AB half AB minus DC this is how you have to okay solve the next question one more question I will give you from this topic and then we will move to next topic so suppose ABCD is a parallelogram and this is not the diagram so ABCD I am saying that ABCD is a parallelogram now there is a point E here E and there is a point F here so E and F are midpoints of AB and CD proved that line segments so this is E this is F so AF this is AF and CE so this is CE trisect diagonal BD so I am making diagonal BD so suppose these points are P and Q so you have to prove that BP is equal to PQ is equal to QB use the properties of parallelogram try to prove that ACF is a parallelogram as soon as you are done let me know try to prove that ACF is a parallelogram and then apply the theorems of triangle should I solve or Dhruti is saying she has done it should I solve guys okay fine two people have done it which topic should I solve next just write down in the comment box I have done quadrilaterals for an hour now so I like to do a different topic okay let me solve okay three people have done it let me solve it now what I need to do as I told I need to prove that ACF is a parallelogram how do I prove it I know that ABCD is a parallelogram so AB is equal to CD opposite sides are equal as E and F are midpoint of AB and CD so half of AB would also be equal to half of CD so I know that half of AB is E and half of CD is CF so AE is equal to CF now I also know that AB is parallel to CD because it's a parallelogram opposite sides are parallel so part of AB which is AB will be parallel to CF also so AE is equal to CF and AE is parallel to CF it gives me a case that AE CF is a parallelogram if it is a parallelogram I can say that now the other part AF is other sides AF would be parallel to EC or part of AF which is QF would be parallel to CP remember this why I am using this because there is a theorem in triangle that if a line drawn from midpoint of a side to opposite side is parallel to third side then that line would be bisecting the third side bisecting the second side rather what I am trying to say by this theorem is that this F is midpoint if I take triangle DCP so F is midpoint of DC now from F I am drawing a line FQ which is parallel to CP it means that this Q is midpoint of line DP so Q is midpoint of DP which says that DQ is equal to QP similarly I can prove it on this side so how I can prove it on this side so I can say that this part is parallel to this part and I know that AE is midpoint and a line is drawn so P would be I am saying here for this part that P is midpoint of BQ similarly it means that DP is equal to PQ so I have proved that DP is equal to PQ and DP is equal to PQ so all three are equal which was the objective or which I needed to prove ok so this is how we need to prove it let's solve ok I will give you one minute to copy it down ok so we will solve as you guys are telling triangular inequalities I will give you a few questions from triangular inequalities so let me search a few questions from triangular inequalities question is show that the sum of the three altitudes triangle is less than the sum of the three sides just prove it should I solve it should I solve ok two people have already done it ok let me solve it what I have to prove so let me draw a triangle for you ABC and let me draw altitudes this is AD this is BE and this is CF so this is going here so all these are altitudes I have to prove that AB plus BC plus CA is greater than AD plus BE plus CF now what I know is that if I take triangle ADB so there AB would be greater than AD on the other side in triangle ADC also AC would be greater than AD so AB plus AC would be greater than 2AD I am just adding these two equations now I am taking triangle BEC and triangle BEA so in triangle BEC why they are because they are hypotenuse both of them are hypotenuse so they would be greater than perpendicular so here BC would be greater than BE and AB would be greater than BE so AB plus BC greater than 2BE and the third one is let me take triangle BFC and triangle AFC so there BC would be greater than CF and AC would be greater than CF so BC plus AC would be greater than 2CF now suppose this is equation 3 this is equation this is 4 this is 5 this is 6 this is 7 this is 8 this is 9 add equation 3, 6 and 9 so 2 times AB plus BC plus CA is greater than 2 times AD plus BE plus CF this 2 to cancelled you get your result ok let me solve next question again I am giving you similar kind of question prove that perimeter of a triangle is greater than some of its three medians just prove it done this is extension of the idea that or concept that some of the two sides is greater than its greater than 2 times the median ok one person has done it 2 more minutes to solve this ok so idea over here is suppose this is ABC and these are the medians so this is D this is E this is F so idea over here is AB plus AC they are the two sides which are converging on A or which are coming out from A that the sum of these two sides would be greater than the median which is being drawn from this vertex which is 2AB similarly BC plus AB would be greater than 2BE and similarly CA plus BC would be greater than 2CF if you add it you get this answer next question next question is so this is some kind of quadrilateral PQRS and there is a point O here where the diagonal coincide you have to prove that PQ plus QR plus RS plus SP is greater than PR plus SQ this is first theorem that you have to prove and the second theorem that you have to prove is the same thing PQ is less than 2 times PR plus SQ let's try to prove it solve ok take your time and solve it how many of you have solved it none of you have solved it till now ok 2-3 more minutes ok I will solve nobody is solving then I will solve so in triangle PQR let's take triangle PQR so in triangle PQR PQ plus QR would be greater than PR some of the two sides is greater than the third side so now let's take triangle RSP the other side triangle RSP so in triangle RSP which are the sides PS plus SR is greater than PR now let's triangle SRQ so SR plus RQ is greater than QR now let's take triangle PSQ so here PQ plus PS is greater than SQ now add all of them you will find that 2 times PQ plus QR plus RS plus SP is greater than 2 times PR plus QS so this 2 and 2 gone you will find out your answer now next thing that I need to prove is so you look at here now let's take triangles these triangles smaller triangle so let's take triangle OPQ so in triangle OPQ OP plus OQ is greater than PQ in triangle ORQ OQ plus OR is greater than QR now let's take triangle ORS their OR plus OS is greater than SR and now let's take triangle OPS in which OP plus OS is greater than SP add all of these you will find that OP plus OP is 2 OP OR plus OR is 2 OR OS plus OS is 2 OS and OQ plus OQ is 2 OQ is greater than this PQ plus QR plus RS plus SP so OP plus OR is PR so 2 times PR and OS plus OQ is SQ so 2 times SQ is greater than this which I needed to prove so this is how you have to solve this now one last question and then we will wrap up our session the question is once again just give me guys let me search a good question so I am drawing a figure the figure is something like this this is a line L this is a point Q this is a point R and this is a point A this is let's draw a perpendicular from here this is AP and this is angular bisector AS of angle S so I am saying AP is perpendicular to L and I am also giving you information that PR is greater than PQ you have to show that AR would be greater than AQ how many of you have done it? ok Sardha I will explain you second part later solve it quickly 2-3 more minutes done how will you say that Sardha you cannot say like that you have to show that which angle is greater angle so if you want to show AR is opposite to greater angle which angle is greater you will have to show that that angle is greater than any other angle and then you will have to show that AR is greater than AQ should I solve this question? ok let me solve so this construction has been done in such a way that PQ is equal to PS so let me take triangle APQ and triangle APS so in this scenario AP is equal to AP PQ is equal to PS because I have constructed it like that and angle P on both sides are equal because that is 90 degree hence this and this and this so side angle side congruence is there it means that AQ would be equal to AS by CPCT now if I know that AQ is equal to AS if this is angle 1 and this is angle 2 so angle 1 would be equal to angle 2 now let's take triangle let's take this triangle triangle ARS and suppose this is triangle P now if PR is greater than PQ so try to understand in triangle ARS angle 2 would be greater than angle 3 because exterior angle is greater than like this 2 would be equal to this angle plus angle 3 I am not adding this angle so this 2 would be equal to suppose this is angle 4 so angle 2 would be equal to angle 4 plus angle 3 so if angle 4 I am not taking then angle 2 would be greater than angle 3 that's what I am trying to write exterior angle is greater than one of the non-adjacent interior angles so angle 2 is greater than angle 3 so this has been done now angle 2 is equal to angle 1 so I can write that angle 1 is greater than angle 3 opposite to angle 1 is AR so angle AR opposite to angle 3 is angle AQ so sorry side AQ so I can write that AR is greater than AQ so I hope you all understood it the meaning of this line PR is greater than PQ is that on PR there would be a part which would be equal to PQ so that is why I have identified a part PS on PR which is equal to PQ so that is why this point PS has been taken on PR so that this PS becomes equal to PQ apart from this there is no meaning of this so as soon as you drop PQ is equal to PS and you prove that angle 1 is equal to angle 2 so then you should go to angle ARS and triangle ARS and you should say that angle 2 is greater than angle 3 and 1 and 2 have already proved equal so 1 would be greater than 3 opposite to 1 AR is there opposite to 3 AQ is there so AR is greater than AQ so I hope you all understood it now it's already time but one question one of you did not understand so I will explain it once more so this was PQRS and there was a point O here so I told that 2 times PR plus SQ is greater than PQ plus QR plus SR plus PS so how will you do you will take this triangle 1 2 3 4 so OR in triangle 1 OR sorry OP plus OQ is greater than PQ so this is first equation in triangle 2 OQ plus OR is greater than QR this is 2 triangle 3 OR plus OS is greater than SR this is third equation and in triangle 4 OS plus OP is greater than PS now what happens you should understand that OP plus OR is equal to PR and OS plus OQ is equal to SQ so if you add OP plus OR so see here OP and OR will give you PR but it is OP and OR is coming 2 times so it will give you 2 PR on this side and similarly OS and OQ will give you SQ but it is coming 2 times so it will give you 2 SQ and that would be equal to greater than this summation so 2 times PR plus SQ is greater than PQ plus QR plus SR plus PS this is it so I hope you all understood it so thank you for joining the session now I am wrapping up the session thank you so much