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In a circle with centre ‘O’, \(\overline {AB}\) is a chord and M is its midpoint. Now prove that \(\overline {OM}\) is perpendicular to AB. 

(Hint : Join OA and OB consider tri-angles OAM and OBM)

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‘O’ is the centre of the circle. 

AB is a chord, M is its midpoint. 

Join A, B to ’O’. 

Now in ΔOMA and ΔOMB 

OA = OB (radii) 

OM = OM (common) 

MA = MB (given) 

∴ ΔOMA s ΔOMB (SSS congruence) 

∴ ∠OMA = ∠OMB (C.P.C.T) 

But ∠OMA and ∠OMB are linear pair 

∴∠OMA = ∠OMB = 90° 

i.e., OM ⊥ AB.

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