Question

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Answer 1

Given – In ∆ABC, AB = AC and a circle with AB as diameter is drawn

Which intersects the side BC and D.

To prove – D is the mid point of BC

Construction – Join AD.

Proof – ∠1 = 90°   ...[Angle in a semi circle]

But ∠1 + ∠2 = 180°    ...[Linear pair]

∴ ∠2 = 90°

Now in right ∆ABD and ∆ACD,

Hyp. AB = Hyp. AC  ...[Given]

Side AD = AD   ...[Common]

∴ By the right Angle – Hypotenuse – side criterion of congruence, we have

ΔABD ≅ ∆ACD    ...[RHS criterion of congruence]

The corresponding parts of the congruent triangle are congruent.

∴ BD = DC    ...[c.p.c.t]

Hence D is the mid point of BC.