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Question
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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Solution
Consider a ΔABC.
Two circles are drawn while taking AB and AC as the diameter.
Let them intersect each other at D, and let D not lie on BC.
Join AD.
∠ADB = 90°...(Angle subtended by semi-circle)
∠ADC = 90° ...(Angle subtended by semi-circle)
∠BDC = ∠ADB + ∠ADC = 90° + 90° = 180°
Therefore, BDC is a straight line, and hence, our assumption was wrong.
Thus, point D lies on the third side BC of ΔABC.
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