# 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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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.

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

Adding (i) and (ii), we get

⇒ BDC is a straight line.

∴ D lies on BC

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