 Hey, hello friends, welcome again to another session on gems of geometry and we have again taken up new theorem And we are going to discuss this theorem and then also see how to prove this theorem So the theorem says let o and i be the circum center and in center respectively of a circle with circum radius r and in radius small r if If d be the distance oi that means the distance between the circum center and the in center is small d then We have to prove that d square is equal to r square minus 2 times small r times capital R Okay, so before that we need some prerequisites to prove this theorem and One is what we have already discussed in previous session. We'll do it here once again. So First one is this so I am joining i o and producing it on the both sides Say this point is let's say p and let's say this side is q Okay, so we had learned in the previous sessions that what did we learn we learned that let's say a l a l and Pq pq are two intersecting cards are two Intersecting cards Right intersecting cards and where are they intersect? Where are they intersecting? They're intersecting at point i So I is the point of intersection. Isn't it? I is the point of intersection So point of let me write it also point of Intersection Correct. I is the point of intersection. So hence we know that a i into i l i l will be equal to q i into ip Is it a i into i l is equal to q i into ip? For this theorem, you have to just check the previous Sessions there we have discussed this at length Okay, so it is true. Then if you see q i Clearly op here is capital R if I have to just mention this is capital R and This is D. So you can figure out q i will be equal to so let me write this as this is q i So q i will be clearly capital R minus small d and Ip will be capital R plus small d Right from geometry q i if you see this Q i here Let me do it. Yeah, so q i if you see this is q i guys So q i is clearly this capital R. This is a whole is R R minus this D is R minus D, right? So let me just delete this as well so that there's no confusion and This part is clearly R plus D without any doubt Correct. So hence I can then write this as R square minus D square. So if you see We are getting closer to R square minus D square from here if you see right So this you keep in mind first of all, so this has to be you know kept in mind. So a i into i l is equal to R square minus D square And now what so let us do some Exercise here as well. So hence if you see if this angle is alpha Okay, if this angle is alpha, then clearly this angle is alpha, right? Why is this? Because angle lying in the same segment. So hence I can say angle B a l B B a l is equal to angle B m l Is it a B m l why angle line angle subtended subtended Subtended by Same arc Right on the same segment are equal, right? Now if you see I a happens to be the internal bisector of B a c because I is the in-center. So I a is the internal bisector So this angle is also alpha is it right and if that angle is alpha by the same logic this By the same logic we will have this angle is alpha, right? Which logic angles up in it by same mark on the same segment I hope the diagram is clear. Similarly. I be is the angle bisector. So if this is beta This is beta Correct, let me draw. Yeah, so this is beta. So this is 2 Right, so and this is clear to everyone. So this is so the angles and the you know the Which angle is which which all angles are equal is very very evident over here Now by external angle theorem, what do we learn this angle guys this angle is alpha plus beta Isn't it if you see why because this angle will be Equal to the external some of the external interior External and opposite so hence alpha and beta. So hence this angle is a you know, so let me write it as well So that it becomes a yeah structured proof. So hence I am writing angle B a I I'm writing angle B a I Is equal to angle angle I a C both are equal to alpha Okay, and similarly angle Why and the reason is I a is angle bisector I a is angle bisector So hence the two angles have to be equal similarly angle a b I a B I is equal to angle C b I and both are equal to beta because I B is Angle by sector Okay, so this is clear. So hence now if you see Angle What L bi Lb L bi is equal to angle L a L a angle L a Sorry, I have to write angle Angle L a C why the reason is this Correct, so L bi L bi I'm sorry, it's not L bi Lb C my bad. So this is Lb C so angle Lb C Lb C is equal to L a C L a C and both are equal to alpha Okay, therefore now you can say angle B I L B I L is equal to angle Or rather I can directly write alpha plus beta y exterior angle theorem exterior angle Theorem in a in a triangle the exterior angle is equal to sum of the interior opposite angle, right? alpha beta also Also, if you see angle Lb I is also alpha plus beta Okay, that means therefore Triangle L bi is I saw silly strangle I Saw silly strangle Therefore therefore BL is equal to I L Correct BL is equal to I L. This is what we learned now If you see guys L M is the dia because it is passing through Oh, so clearly angle L B M so angle Angle L B M is 90 degrees 90 degrees once again see the diagram L B M is 90 degree or let me do it here Yeah, so hence clearly angle L B M is equal to 90 degrees. Why? L M is dia is the diameter so hence angle subtended by diameter on the circle is 90 always Clearly clear. Also, I have drawn I y is perpendicular to Let's say AC right. This is construction Construction So now what do we get we have in triangle in triangle? L B M Angle L B M is 90 degrees, isn't it? L B M is 90 degree therefore therefore L B L B by L M L B by L M is sine alpha Isn't it? So hence L B is equal to L M sine alpha Right and what is L M guys is the diameter of the circle. So it is 2 R Right R is the circum radius. So 2 R sine alpha Isn't it? O P is equal to O L is equal to O M right 2 R sine alpha L V is this now if you see L B was equal to L I so therefore L I L B This side is equal to this side, right? So hence now you can say L V is equal to L I is equal to 2 R sine alpha Okay, friends now consider in triangle I y a I y a angle y is 90 degrees Right. So it's again. I can say Sine alpha in this case is equal to small r I Y divided by I a Okay, so this implies I a is equal to r upon sine alpha, right? now so we have to Find out L I into I a so L I was 2 R sine alpha into I a is r by sine alpha from here and From here. Okay, so what do we get we get L I? into I a is equal to 2 R into small r and what was li into I a if you see Li into I a here is it it is Li into I a right was r square minus d square so hence combining the two I can say This implies r square minus d square is equal to 2 times r times r Right. Once again, this was li into I a is this r square minus d square here So hence now right so rearranging you can say d square is equal to r square minus twice r times r hence Proved I hope you understood the theorem Okay, so hence any point inside Or rather, you know the distance I o is d square the distance between the circum radius circum center and in center is nothing but Circum radius square minus twice circum radius times in radius. This is what the theorem was All about