 Hello and welcome to another session on gem surf geometry and we are going to take up another theorem here in this session and the theorem says if a cyclic triangle has perpendicular diagonals intersecting at point P, the line through P perpendicular to any side bisects the opposite side. So what we are going to do is we will first construct this diagram and then we will prove the theorem. So let us construct the diagram. So what do we need? It says cyclic quadrangle has perpendicular diagonal. So let us just draw two lines and basically perpendicular line. So let me say this and this is a line and let me draw a perpendicular line to this line. So from here I am taking, yeah okay so I have to draw a perpendicular line from this point. These are the perpendicular lines, right? Now we have to just make sure that there is a quadrangle, a cyclic quadrangle. So and these two lines should be the diagonals or the diagonal should fall on these lines. So let us just draw any the circle. Let us say I am taking this as center and from here passing through any. Yep, so this is my circle and what I am going to do is I am just going to delete these points or other. So what I am going to do is, so I have got a circle where two perpendicular lines are there. So I can now identify these points. Let us say this is point F. So let me rename is rename as P okay P and let us say this point is Q. So let me rename this as Q. So this is Q and I need this D is let me rename it as R okay and let us say this point here. Let us rename it as S PQRS okay and then I can hide these lines and objects which one. So I would not be needing this E, I would not be needing the C neither this line nor this line or this point A nor this point B. Okay now we can draw or join these segments. So let's say QS and PR. So PR and QS these are two diagonals and now we can create a polygon or a cyclic quadrangle required. So P, Q, RS is the required cyclic quadrangle where the diagonals are perpendicular to each other right. And how do I know it? You can coin this or let this name be T okay. So you can check this angle will be ST 90 degrees right. So 90 degrees and let me see if I can move this point around. No but then the 90 degrees thing is gone. So this is the only thing okay I can do like this yeah. So if I maintain point R to be on this line then 90 degrees fair enough. So this is good enough construction. I don't need this. So let me just delete that part okay. So this is the required what do you say the diagram. Now let me yeah this is perfect enough. Now what I am let us go to the second part. So it says the line through P okay. So this name has to be P but okay so we have to rename it. So P, Q, RS, T. So let me rename this as T okay. And so that I can name this as P okay. So let me rename this as P so that we do not different or we do not deviate from whatever the question is given now or the theorem is. So now it says the line through P perpendicular to any side. So let's pick any side arbitrarily. Shall we pick RS? Let's pick RS and let's draw up a perpendicular line on RS. So passing through P that is okay. So here is what I have done and now this is the point of intersection of. So let us call this name as U right okay perfect. So this is what now the question is or they are saying that U will be the midpoint of Q, T is it. So first of all let's try to measure it is it so so angle distance right. So I am measuring T, U it comes out to be 3.9 and this one oh wow so it is actually midpoint. So U is the midpoint of T, Q and let me just change the circle a bit. Yeah but there also you can see the relationship is intact that is that is the midpoint actually and now 2.6 here also till I am maintaining the fact that they are 90 degrees. So this is see till I maintain 90 degrees I will get the same right 3 this is going to be same. So hence till the diagonals are perpendicular to each other our job or whatever they are saying is true. So let's try to prove it now. How do we prove it? So let me delete this I will not require this part full. So let me delete this as well oh sorry no I need not delete that. I have to simply switch off this line because this is not needed okay and then take a segment and join UP right. Oh in fact I had to do I had to do the other part as well so sorry. So we what we have to do is we have to let it be and this point is needed yes and now I can hide this line that's not needed anymore. So I am hiding it and I am joining UD or let me name this rename it as UV right. So this looks good all towards the end of the English alphabet very good. Now I have to simply join U and V okay and now I can also show this that this angle is 90 which one. So if I measure this angle S, V and P perfect I don't need this value so let me just take this value off okay. So this is the point V let me highlight this point V here oh this is V this is alpha right. I don't need this either so let me turn this off as well okay. Can I name and value, value, caption oh show label so let me just take it away. Yep this looks good this is the diagram this is the construction and now I need to prove it so how do we prove it let's try to prove this theorem. So what are we going to do so okay let's see what all is given to us okay. Now let us say this angle is X okay if this angle is X then okay first of all this is 90 degrees so this angle is 90 minus X no doubt about it right and if that is 90 minus X then this angle is X right. So what I am saying is angle V PS will be 90 minus X and angle RPV RPV is equal to 90 minus 90 minus X hence X so this should be obvious right okay looks good now if this is X then this X X so I am writing angle UPT is X no problem and also this angle is X right. So I am writing angle QPR or UTP or first let me write QPR is equal to angle Q SR why is that because this is angle in the same segment angles in the same segment in the same segment of a circle are equal is it. So hence it is subtended by both QR or basically QR is the you see this is the chord it subtends an angle at T and the same chord subtends an angle at S they are in the same segment so they must be equal and this is equal to X now in triangle this is a very important finding so in triangle UPT what do we get we get UP will be equal to UT is it UP is equal to UT UP is equal to UT right why because this is isosceles triangle now angles opposite angles are equal to see X is equal to X so isosceles triangle no problem now similarly you can say that UP is equal to QP so instead of saying similarly let us try and prove it okay so okay so what and how can we prove that so clearly this angle is 90 minus X is it this is 90 minus X without doubt okay so yeah so this angle is how much this angle will be what will be this angle this angle is so this is X this angle is X clearly this is 90 degree isn't it so this angle has to be 90 minus X now I hope this is clear so what I'm saying is angle T our angle QPU is equal to 90 minus X vertically opposite angle vertically opposite angle isn't it QPU QPU is 90 minus X also in triangle QTP QTP 90 degrees plus X plus angle TQP TQP is equal to 180 degrees by ASP angle some property of a triangle so you can find out TQP is equal to 90 degrees minus X right that means I'm writing here now in triangle UQP UQP angle UQP is equal to angle UPQ or QPU whichever way 90 minus X right 90 minus let me write here 90 minus X okay so hence triangle UQP is a nice also this triangle hence hence hence UQ is equal to UP correct now let us say this was 1 and this is 2 so you can say from 1 and 2 what do we get UP is equal to UT is equal to UP is equal to UT is equal to UP hence U is the midpoint U is the midpoint of side QT right hence proved this is what we wanted to prove so again just to reiterate what word did the theorem say it said that if you have a quadrangle cyclic quadrangle such that the diagonals are perpendicular to each other then if you draw a line perpendicular to one of the sides passing through the point of intersection P then it is going to cut the opposite side exactly at the midpoint once again there are two diagonals of a cyclic quadrangle which are perpendicular now if you draw a line such that the line segment is perpendicular to one of the sides like this and passes through that point of intersection of the diagonals then it will go and bisect the opposite side QT and this will be true for any pair any side for that matter so if you draw up a perpendicular here and take it like that so if this is 90 degrees then this will be the midpoint this will be the midpoint like that so it will be holding true for any side right so this is what the theorem was saying and we constructed the diagram and we proved it as well so I hope you understood understood this theorem and it's proved as well thanks for watching bye-bye see you in the next video