Question

Prove that the circle drawn with any equal side of an isosceles triangle as diameter bisects the base

Answer 1

A $△ABC$ in which AB=AC and A circle is drawn by taking AB as diameter which intersects the triangle at D join AD
IN $△ADB$ and $△ADC$ we are given AB=AC
$AD=AD$
$\angle ADB={90}^{\circ }$ (Angle in a semi-circle)
$\therefore \angle ADC={90}^{\circ }$
$\therefore \angle ADB=\angle ADC{90}^{\circ }$
Hence $△ADB\cong △ADC$
(RHS Theorem of congruence )
$BD=DC$ [ corresponding sides of congruent triangles]