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If circles are drawn taking two sides of a triangle as diameters. Prove that the point of intersection of the circle lie on the third side.

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 does 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 ?ABC.


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