 Hello guys welcome again to this session on gems of geometry and today we are going to take up one more interesting theorem and the name also is very interesting the name is butterfly theorem right so by looking at the impression of the image you can say or this figure you can see that you know there is some kind of butterfly which is you know you know getting you know prominently manifested over here so for example what I'm trying to say is just observe the shape admcb right so admcb looks like but during our childhood we used to make some kind of a thread structure right so using a thread we used to make and make a butterfly you know you know some kind of structure with our fingers and threads I don't know if you you have done the same thing in your childhood but it looks similar to that so that's called butterfly theorem and what exactly is the theorem the theorem says that if a chord pq is there on a circle and m being the midpoint of that chord then if you let two more chords pass through m let's say ab and cd are the two extra chords which are passing through m and you join adnbc like that so adnbc are joined and let the point of intersection of pq and adbx and that of pq and bcby okay so these are two point of intersections now what the theorem suggests is m also would be the midpoint of xy okay so xy is this segment and m also will be the midpoint of xy right so pq is a chord m is the midpoint you are letting two chords pass through m ab and cd find the point of intersection of ad and pq call it x find the point of intersection of bc and pq call it y and you will be surprised to see that m is the midpoint of xy as well that is x and y are equidistant come what may as you know whatever be the configuration so just to validate let's see whether that is true so what we're going to do is we are going to change the configuration of this diagram and let's see for different positions of all these chords and points whether actually the value or the theorem is valid so let me change the position of the points now and keep an eye on my and mx and you know you can also see that pm and qm would be same because m happens to be the midpoint right so here is what I am doing is I am changing the position of the point on circle right and I am also going to change this configuration so that you can see that at whatever be the configuration m is the midpoint of xy right so it always so it's coming closer you can see that it's coming closer and eventually it will merge so hence xy uh m is midpoint of xy right so this is kind of validated through demonstration what remains to be done is to to be you know let's try and prove this analytically so that's what we are going to do in the next part of the session so let's go to proving so now having verified the butterfly theorem through the software it's now remaining to prove the theorem itself correct so I have stated the theorem first and I have done some construction which I'll anyways be explaining in some time and then we'll go for the proof okay so the statement of the butterfly theorem says through the midpoint m of a chord pq of a circle any other chords ab and cd are drawn chords ad and bc meet pq at point x okay and y then m is the midpoint of xy so we have to prove that m is the midpoint of this xy here m is the midpoint of xy okay so let's try and prove before that we have to do some prerequisites so that it becomes easier for those who are not initiated into proving or let's say similar triangles and other things so we will do some kind of basic work groundwork preparation and then we'll go for the proof so for doing this proof there is some construction uh so I will list that down so construction is this so what all constructions have been done so I've drawn x e perpendicular to ab and xf xf is perpendicular to cd okay similarly yg is perpendicular to cd and yh is perpendicular to ab right this is the construction now we have to be very clear with similar triangles here so we are we are going to use a lot of similarity principles so hence let's try first of all uh proving that triangle so these are similar triangles pair of similar triangles I'm going to write so please be you know careful and keep that in your mind so triangle xfm is similar to triangle um ygm ygm I hope there is no problem in this and why is this because one is this vertically opposite angle and then other one is this 90 degree so they are by a similarity right so by a similarity criterion right so this is one yeah so please bear in mind let me just take away these okay so you know similarly you can you can prove that this is point number one point number two triangle uh similarly xem xem is now similar to triangle y and uh yhm correct same same logic a similarity so I'm not going to explain you can stop or pause the vv video and then prove for yourself okay now third one also is this if you see triangle xae is similar to triangle uh ycg ycg and why is that very clearly if you see this angle a is equal to angle c angles in the same segment okay I'm angle subtended by an arc in the same segment is equal so that ways and there are two right angles again so by a similarity they are also similar so I can write that as a similarity here also right very good now one more similar triangle set I can write and that is xdf triangle xdf and triangle ybh ybh and again by same a similarity I hope I have written the orders of the vertices in the correct order right so xdf yes and ybh very good so why because uh again there these are those two angles in the same segment by same arc and 90 degrees so hence no doubts about it that these two triangles are similar right so for neatness we'll just remove these marks very good now if you see I have uh kept or I've called pm pm you can sorry xm can you see these are some things which we have made xm is equal to x and ym I have written as y for convenience and pran b our self-explanatory p is xe r is xf b is yh and a is gy right so this is clear if this is clear let's now go for the proof how to uh go for the proof so let me now write it here now since the triangles are similar can you not say that x upon y and let's take the triangle xem so this triangle and this triangle first take these two triangles so x upon y will be equal to p upon b isn't it similarly let it be one similarly let me just remove all of this yes and this was y this is y okay now let's take the other two triangles which one these these two triangles here also x upon y can be written and what is that my dear friends so x upon y is equal to um p upon a no x upon y is r upon a sorry r upon a right by similarity also okay very good now um if you see we multiply these equations together so what will you get x square by y square is equal to p upon b into r upon a which is equal to p upon a into r upon b and why did i do this you'll get to know this why later okay p by a what's p by a guys if you see can i say p by a is equal to p by a look at the figure now so this is p this is a so p by a will be this side upon that side isn't it similar triangles we just proved so x a upon y c and similarly r by b if you look carefully it will be dx or xd by y b isn't it now if you remember the theorem where if you have a circle and there are two chords like that intersecting at point let's say m and let's say this is ab this is cd so we know that am into mb is equal to dm into mc correct this is the theorem which we have proved proven earlier so you can check this in the previous videos right so we're using this theorem can i not say x a by this top one x a by xd should be x a so look at the figure where is x here is x where is a here is a and where is d here is d so ad is the chord here in isn't it ad is the chord and x is the point of intersection so can i not write that as px into x q okay just to make it more clear i can write px into qx so this is into sign okay so let me just demarcate it like that so px into qx px into qx is equal to x a into xd by this theorem similarly in the denominator yc yb check yc yb so here is b here is y here is c so bc is the chord and pq is the intersecting chord so you can again write qy qy into yp yp okay so now what is px guys if you see px is nothing but pm minus x pm minus x look carefully and what is qx qx qx is qm into qm sorry qm plus x am i right so qx is qm plus x yes because mx is small x now qy where is q so here is q look at this here is q qy will be simply mq minus m y isn't it so mq minus m y what can i write that as so qm minus x again what sorry minus y minus y i can write that as qm minus y and q q next one is yp right this yp yp is nothing but ym plus mp so ym ym is y small y plus mp so pm like that and we know that m is the midpoint of pq so clearly pm is equal to mq this was the starting point guys pm is equal to mq so what would this relation become this will simply become pm and let it be let's say small s okay for convenience so that we can write now so this is s minus x times s plus x divided by s minus y s plus y okay so and where did we start from x square y square right so this relation is basically nothing but x square by y square is equal to s square minus x square and s square minus y square s minus x s plus x is s square minus x square so now from here cross multiplication x square s square minus x square y square is equal to sorry this is y square guys so y square s square minus x square y square correct yeah and now if you see this can be eliminated and since s is a non-zero quantity what is s guys s was pm isn't it pm it's a non-zero quantity so you can eliminate s square from both sides as well so you get x square is equal to y square so you get x square is equal to y square and that means x is equal to y right this is what we needed to prove x equals to y that means what if x equals to y then we can say x see x is equal to y prove that x is equal to y when is that possible when m is the midpoint of x y so hence you can say m is the the midpoint of x y right hence okay i hope you understood the proof this is what is called butterfly theorem folks thank you