Question

please help me to solve the problem​

Answer 1

Answer:

I don't know this answer

Answer 2

Answer:

Theorem 10.3

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The perpendicular from the center of a circle to a chord bisects the chord.

Given : C is a circle with center at O. AB is a chord such that OX 1 AB

To Prove : AC bisect chord AB i.e.

AX = BX

Proof :

In AOAX & AOBX

ZOXA = ZOXB

OA = OB

OX = OX

:: AOAX = AOBX

AX = BX

Hence, Proved.

A

X

B

(Both 90°, given)

(Both Radius)

(Common)

(RHS Rule)

(CPCT)