 Hello and welcome to the session. In this session we will discuss the following question which says ABC is an isosceles triangle in which AB is equal to AC. Circumstriped above the circle show that BC is bisected at the point of contact. This is the figure in which we have triangle ABC. Circumstriped about a circle with center O. We know that the lengths of tangents drawn from an external point to a circle are equal. This is the key idea for this question. Now let's move on to the solution. Now let's see what is given to us in the question. We have that ABC is an isosceles triangle that if we have AB is equal to AC and we have to show that BD is equal to DC. Now let's see its proof. Now as you can see in this figure AF and AE are the tangents from external point A to the circle. So we have AF would be equal to AE since they are the tangents from the point A and we know that the lengths of the tangents drawn from an external point to a circle are equal. So next we have BD and BF are also the tangents from point B. So we have BF is equal to BD since they are the tangents from the point B. In the same way we have CD is equal to CE since they are also the tangents which are drawn from the point C. We take this as equation one, this as equation two and this as equation three. Now next we have adding equation one, two and equation three, we get AF plus BF plus CD is equal to AE plus BD plus CE. Now this implies, now if you look at the figure you can see that AF plus BF is equal to AB. So we have here AB plus CD equal to AE plus CE, this is equal to AC. So we have AC plus BD. Now this further implies that CD is equal to BD since we have given that AB is equal to AC. That is we get that BD is equal to DC. So we have proved this. So final answer is that BC is bisected at the point of contact. So this completes the session. Hope you have understood the solution for this question.