Question

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

Answer 1

Given: Angle bisector of angle A and perpendicular bisector of BC intersect.

Let the bisector of A meet the circumcircle at E.

In $△ABE$ and $△AEC$,

$\angle BAE=\angle CAE$ {AE Angle bisector of A}

$\therefore BE=EC$ .....(1) {converse of the angle subtended by the chord of equal length at the same point are equal}

Let D be the midpoint of BC and now connect ED.

In $△BDE$ and $△CDE$,
$BE=CE$ ....from eq. (1)
$BD=CD$ ...{$D$ is mid point of $BC$}

$DE=DE$ ...common side

$\therefore △BDE\cong △CDE$ ...SSS test of congruence

$\Rightarrow \angle BDE=\angle CDE$ .....c.p.c.t.

Also, $\angle BDE+\angle CDE={180}^{\circ }$ ...angles in linear pair

$\therefore \angle BDE=\angle CDE={90}^{\circ }$

Therefore, $DE$ is the right angled bisector of $BC$.

Hence, they intersect on the circumcircle of triangle ABC