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: ΔABC is inscribed in a circle. Bisector of ∠A and perpendicular bisector of BC intersect at point Q.

To prove: A, B, Q and C are con-cyclic.

Construction: Join BQ and QC.

Proof: We have assumed that, Q lies outside the circle.

In ΔBMQ and ΔCMQ,

BM = CM  ...[QM is the perpendicular bisector of BC]

∠BMQ = ∠CMQ   ...[Each 90°]

MQ = MQ   ...[Common side]

∴ ΔBMQ ≅ ΔCMQ  ...[By SAS congruence rule]

∴ BQ = CQ  [By CPCT] ...(i)

Also, ∠BAQ = ∠CAQ   [Given] ...(ii)

From equations (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 con-cyclic.

Hence proved.