Question

Question 5
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 $\Delta ABC$ is inscribed in a circle.
The bisector of $\angle A$ and the perpendicular bisector of BC intersect at point Q.


To prove that A, B, Q and C are concyclic.
Construction : Join BQ and QC

Proof
We have assumed that Q lies outside the circle.
In $\Delta BMQand\Delta CMQ$
BM = CM [QM is the perpendicular bisector of BC]
$\angle BMQ=\angle CMQ\left[each{90}^{\circ }\right]$
MQ = MQ [Common side]
$\therefore \Delta BMQ\cong \Delta CMQ$ [by SAS congruence rule]
$\therefore$ BQ = CQ [by CPCT] … (i)
Also, $\angle BAQ=\angle CAQ$ [given] ..........(ii)
From Eqs. (i) and (ii) , we can say that Q lies on the circle
[equal chords of a circle subtend equal angles at the circumference]
Hence, A, B, Q and C are concyclic.