Question

to prove that A perpendicular drawn from the center of a circleon it's chord bisect the chord​

Answer 1

Answer:

To prove that the perpendicular from the centre to a chord bisect the chord.

Consider a circle with centre at O and AB is a chord such that OX perpendicular to AB

To prove that AX=BX

In ΔOAX and ΔOBX

∠OXA=∠OXB [both are 90 ]

OA=OB (Both are radius of circle )

OX=OX (common side )

ΔOAX≅ΔOBX

AX=BX (by property of congruent triangles )

hence proved.

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