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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.
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Solution
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.
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