# In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC

**Solution:**

Angles in the same segment are equal.

∠BAE = ∠CAE {Angles inscribed in the same arc}

arc BE = arc CE

Hence chord BE ≅ chord CE …(1)

In △BDE and △CDE,

BE = CE {from (1)}

BD = CD {ED is perpendicular bisector of BC}

DE = DE {common side}

Therefore, △BDE ≅ △CDE {By SSS criteria of congruence}

∠BDE = ∠CDE {By CPCT}

Now, ∠BDE + ∠CDE = 180º {linear pair}

Hence ∠BDE = ∠CDE = 90º

Thus, DE is the perpendicular bisector of BC.

Therefore, the perpendicular bisector of side BC and the angle bisector of ∠A meet on the circumcircle of triangle ABC

**Video Solution:**

## In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC

### Maths NCERT Solutions Class 9 - Chapter 10 Exercise 10.6 Question 10:

**Summary:**

It is given that in any triangle ABC, the angle bisector of ∠A and perpendicular bisector of BC intersect. We have proved that they intersect on the circumcircle of the triangle ABC.