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.

Answer 1

Data: Two circles are drawn taking PQ and PR of a triangle as diameter. 

Let these intersect at P and S. 

To Prove: The point of intersection ‘S’ is on the third side QR of ∆PQR. 

Construction: Join PS. Proof: QAP is a diameter. 

∴ ∠QSP = 90° (angle in the semi circle) 

Similarly, ABR is a diameter. 

∠PSR – 90° (angle in the semicircle) 

∠QSR = ∠QSP + ∠RSP = 90 + 90 

∠QSR = 180° 

∴ ∠QSR is straight angle. 

∴ QSR is a straight line. 

∴ Point ‘S’ is on third side QR of ∆PQR.