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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)
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