If the 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.
Asked by bansari Butani
| 19th Feb, 2014,
02:14: PM
Expert Answer:
Dear Student,
Consider a
ABC.
Two circles are drawn while taking AB and AC as the diameter.
Let they 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 third side BC of ΔABC.
Thanks and Regards
Toppers Team


∠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 third side BC of ΔABC.

Thanks and Regards
Toppers Team
Answered by
| 19th Feb, 2014,
06:19: PM
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