Question

Prove that the median to the base of an isosceles triangle is perpendicular to the base.

Answer 1

Let ABC is an isosceles triangle with AB=AC and let AD be the median to the base BC.then D is the mid-point of BC.

$\Delta ABD\cong \Delta ADC$ by SSS

so, $\angle ADC=\angle ADB=x$

$\therefore \angle ADB+\angle ADC={180}^{\circ }$

$x+x=180$

$x=\frac{180}{2}={90}^{\circ }$

$\therefore \angle ADB=\angle ADC={90}^{\circ }$