Question

two circles circles are drawn taking two side of a triangle as diameter prove that the point of intersection of the circles lie on the third side​

Answer 1

Step-by-step explanation:

Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D.

To prove: D lies on BC

Construction: Join A and D

Proof: ∠ADB = 90° (Angle in the semi-circle) ...(i)

and ∠ADC = 90° (Angle in the semi-circle) ...(ii)

Adding (i) and (ii), we get

∠ADB + ∠ADC = 90° + 90°

=> ∠ADB + ∠ADC = 180°

=> BDC is a straight line.

Hence, D lies On third side BC.

Answer 2

Answer:

here it is

Step-by-step explanation:

Proof: ∠ADB = 90° (Angle in the semi-circle) ...(i)

and ∠ADC = 90° (Angle in the semi-circle) ...(ii)

Adding (i) and (ii), we get

∠ADB + ∠ADC = 90° + 90°

=> ∠ADB + ∠ADC = 180°

=> BDC is a straight line.

Hence, D lies On third side BC.