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: $△ABC$ is an isosceles triangle with AB = AC. A circle is drawn taking AB as diameter which intersects the side BC at D

To prove: BD = DC

Construction: Join AD

Proof: $\angle ADB={90}^o$ [Angle in a semi-circle is right angle]

$\angle ADB+\angle ADC={180}^o$

$\Rightarrow \angle ADC={90}^o$

In $△ABD$ and $△ACD$

$AB=AC$ [Given]

$\angle ADB=\angle ADC$

$AD=AD$ [Common]

$\therefore △ABD\cong △ACD$ [RHS congruence criterion]

So,

$BD=DC$ [By CPCT]