 Hi friends welcome back to the sessions on gems of geometry and in this session we are going to take up a series of theorems which are related to circles. In the previous series we saw lots of theorems related to a triangle. We saw middle triangle or thick triangle and many other properties of those triangles and from this session onwards we are going to deal with lot of theorems which are related to circles and yes on and off we will also see mixed concept of circles and other polygons. So the first session is about two intersecting cards and it is a very famous and very common result as well. So in this session we are going to discuss this theorem which is that if two cards B E and C D intersect at a point F then B F into F E is equal to C F into F D. So you can see in this diagram F is the point of intersection of C D and B E. They are intersecting at point F. Now it is given that B F into F E or we have to basically prove that B F this B F this part into F E the other part of the same card is equal to product of the two parts of the other card as well. So let's see whether it is true. So we will first demonstrate that yes indeed it is true and then we will try to prove it. So here is a circle four points you can see C B D and E and I am going to change the positions of D to change the positions of other points as well especially F. So let's see if the result is always same. So here I am changing the location of point D. Okay so can you see I am changing the point D. So you can see the result you can see here both the product LHS of this relationship and RHS of this relationship are equal. Can you see that? So it's always same. So hence it looks like it looks like the theorem is true right valid. So I am changing E as well so that there is no confusion C E as well. So everything every time I change the location of E that means if you can see the location of F is changing and hence the four segment lengths are changing and you can see the value remains the same. Let me try at this value as well. So yeah there is absolutely no change wherever I take the points and hence now if you see if I change the radius of the circle will that also impact or will there be no impact if you see I am changing the radius and still there is no difference guys. Yeah so they are even if the radius is changing or the location of B any four points you are changing even if the radius changes doesn't matter. So if you see the LHS and RHS of the given result always are equal and you can see that by this demonstration we can show at least that the these products are equal and here is the proof as well. So you can a very simple proof based on similarity of triangles. So you join B D and C E so in triangle B F D here B F D and C F E if you see B F D and C F E these two triangles are similar I can prove that how so for example B F D this angle this angle is 37.61 it is demonstrated here and this angle is also 37.61 so both are equal has to be why because they are vertically opposite angle so whatever be the case if you see the vertically opposite angles are always same right and then B F D so the other triangle is B D F so B D F is another angle and C E F these two sorry C F F it should be C E F yeah C E F and B D F both are same why because you can you can check here this is alpha 56.74 beta 56.74 and irrespective of the point E the two angles are not going to change can you see if it is 56.74 in this case correct so hence these two angles are all also equals so these two triangles this one B F D and C F E are always similar by double A or AA similarity criteria okay so we know that if two triangles are similar their sides are proportional so hence I can write F D upon F E is equal to F B upon F C repeat F D upon F E these are the corresponding sides so hence F B upon F C corresponding sides ratios are equal so I can write that and then you cross multiply to get the desired result B F into F E is equal to C F into F D so what did we learn guys we learned that if two chords you know intersect each other so then their product of their segments are equal I can take a special case when this is let's say the dia okay can you see it's diameter now both are intersecting at the center almost so you know so the lengths are almost six point almost six isn't it so yes so this is yeah yes six all are six all the four parts now are equal to six can you see so hence two intersecting diameters anyways we'll divide each other in you know two radii so then also the the theorem will be valid isn't it so that's the learning of this theorem