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If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side

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Given : Sides AB and AC of a triangle ABC are diameters of two circles which intersect at D. 

To Prove : D lies on BC. 

Proof : Join AD 

∠ADB = 90° ...(i) [Angle in a semicircle] 

Also, ∠ADC = 90° ..(ii) 

Adding (i) and (ii), we get 

∠ADB + ∠ADC = 90° + 90° 

⇒ ∠ADB + ∠ADC = 180° 

⇒ BDC is a straight line. 

∴ D lies on BC 

Hence, point of intersection of circles lie on the third side BC. Proved.

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