Perpendicular, OM, from centre of the circle, O, to the chord, AB, bisects the chord. To Prove:
Proof:
Join OA and OB
OA= OB radii of the circle
ΔOMA ≅ OMAΔ Hypotenuse Leg Theorem
BM = MA Congruent Triangle
Hence proved Perpendicular from centre of the circle to the chord bisects the chord
Related Theorems
Diameter perpendicular to the chord will bisect the chord Right bisector of the chord will pass through the centre of the circle.
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