Question

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.​

Answer 1

Answer:

BAE=∠CAE ....Angles inscribed in the same arc

∴arc BE=arc CE

∴ 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

∴△BDE≅△CDE ...SSS test of congruence

⇒∠BDE=∠CDE .....c.a.c.t.

Also, ∠BDE+∠CDE=180

∘

...angles in linear pair

∴∠BDE=∠CDE=90

∘

Therefore, DE is the right bisector of BC.

Answer 2

Answer:

I was not able to post the big answer... so I have attached a pic.