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
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.
∠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
Answered by | 19th Feb, 2014, 06:19: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number