If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th
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Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC  Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle) ...(i)   and  ∠ADC = 90° (Angle in the semi-circle) ...(ii)  Adding (i) and (ii), we get  ∠ADB + ∠ADC = 90° + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.
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