Question

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect they will intersect on the circumcircle of the triangle.

Answer 1

Given: $\Delta ABC$ and $l$ perpendicular bisector of $BC$

Refer image,

To prove: Angles bisector of $\angle A$ and perpendicular bisector of $BC$ intersect on the circumcircle of $\Delta ABC$

Proof: Let the angle bisector of $\angle A$ intersect circumcircle of $\Delta ABC$ at .P Join $BP$ and $CP$.

$\Rightarrow \angle BAP=\angle BCP$

[Angles in the sme segment are equal]

$\Rightarrow \angle BAP=\angle BCP=\frac{1}{2}\angle A$.........(1) [$AP$ is bisector of $\angle A$]

Similarly, we have

$\Rightarrow \angle PAC=\angle PBC=\frac{1}{2}\angle A$.........(1) [

From equation (1) and (2), we have

$\Rightarrow \angle BCP=\angle PBC$

$\Rightarrow BP=CP$

[$\because$ If the angles subtended by two Chords of a circle at the centre are equal, the chords are equal]

$\Rightarrow P$ is on perpendicular bisector of $BC$.

Hence, angle bisector of $\angle A$ and perpendicular bisector of $BC$ intersect on the circumcircle of $\Delta ABC$.