Question

. In the adjacent figure AB is a chord of circle and O is centre of the circle. If OM is perpendicular to AB, prove
that
Mis mid-point of AB.
(Hint: Join OA and OB and prove AOAM = AOBM)
please step by step answer​

Answer 1

Answer:

Hi ,

Given : AB is a chord of a circle with center

O . CD is the diameter Perpendicular to AB.

RTP : AD = BD

proof :

Let P is the intersecting point of AB and CD.

From triangles APD and BPD

<APD = <BPD = 90° ( GIVEN )

DP = DP ( Common side )

AP = PB ( diameter bisects chord )

Therefore ,

triangle APD congruent to triangle BPD

( SAS creiteria )

AD = BD ( corrospoding parts of congruent

triangles )

Hence proved.

I hope this helps you.

Answer 2

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