Consider a ΔABC
Two circles are drawn while taking AB and AC as the diameter
∠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
Thus, Point D lies on the third side BC of ΔABC.
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