 Hello and welcome to the session. My name is Mansi and I'm going to help you with the following question. The question says in any triangle ABC if the angle bisector of angle A and perpendicular bisector of BC intersect prove that they intersect on the circum circle of the triangle ABC. So let us start with a solution to this question. First of all let us see what is given to us. We are given a triangle ABC inscribed in a circle with center O. E is a point on the circle such that AE is the internal bisector of angle BAC that means angle BAE is equal to angle CAE. Now we have these two angles are equal and D is the midpoint of BC that means BD is equal to CD. Now what we have to prove here that DE is the right bisector of BC that means angle BDE is equal to angle CDE is equal to 90 degree. So we have to prove that angle BDE is equal to angle CDE is equal to 90 degree. First of all we do some instruction we join B to E and C to E. Now let us start with the proof. First of all we see that in triangle BDE and triangle CDE we have BE is equal to CE because angle BAE is equal to angle CAE since we have seen that AE is the angle bisector. Now these two angles are equal therefore arc BE is equal to arc CE and since these two arcs will be equal so BE will also be equal to CE. Also we have BD is equal to CD because this is the perpendicular bisector this is the bisector. Now DE is equal to D that is a common side in both the triangles. So by SSS criteria we have triangle BDE is congruent to triangle CDE. So let us write down what we have just seen. We see that in triangle BDE and triangle CDE we have BE is equal to CE because angle BAE is equal to angle CAE therefore arc BE is equal to arc CE and this implies that cod BE is equal to cod CE. Also we have BD is equal to CD that is given to us and we have DE is equal to DE because that is a common side in both the triangles. Therefore by SSS criterion of congruence we will have triangle VDE is congruent to triangle CDE. Since these two triangles are congruent so this implies that angle VDE is equal to angle CDE this we get by CPCT that is corresponding parts of congruent triangles. Also we see that angle BDE plus angle CDE will be equal to 180 degree because these are two angles in a linear pair and angles in a linear pair they sum up to 180 degrees. So we have angle BDE plus angle CDE is equal to 180 degree since both these angles are equal. So this implies angle BDE is equal to angle CDE is equal to 90 degree hence DE is the right bisector of VC. So I hope that you understood the question and enjoyed the session have a good day.