Question

prove that the perpendicular from the centre of a circle to a chord bisects 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 OO and ABAB is a chord such that OXOX perpendicular to ABAB

To prove that AX = BXAX=BX

In \Delta OAXΔOAX and \Delta OBXΔOBX

\angle OXA = \angle OXB∠OXA=∠OXB [both are 90 ]

OA = OB OA=OB (Both are radius of circle )

OX = OX OX=OX (common side )

ΔOAX\cong ΔOBXΔOAX≅ΔOBX

AX = BXAX=BX (by property of congruent triangles )

hence proved.

solution

Step-by-step explanation:

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