Question

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.

Answer 1

Given: Circles are described with sides AB and AC of a triangle ABC as diameters. They intersect at a point D. To Prove: D lies on the third side BC of AABC.

Construction: Join AD.

Proof: Circle described on AB as diameter intersects BC in D.

∴ ZADB = 90°

Angle in a semi-circle

But LADB + ZADC = 180°

Linear Pair Axiom

∴ ZADC = 90°.

Hence, the circle described on AC as diameter must pass through D. Thus, the two circles intersect in D.

Now, ZADB + ZADC = 180°.

∴ Points B, D, C are collinear.

∴ D lies on BC.