Question

Perpendicular drawn to the unequal side of an isosceles triangle divides the side intoparts and also divide the whole triangle into twotriangles

Answer 1

In $△ADC\&△ADB$,
$\angle ADC=\angle ADB={90}^{o}AD=ADAB=AC$

Hence, by Hypotenuse Leg congruency,
$△ABD\cong △ACD$

We know Corresponding sides of congruent triangles are equal, hence
$BD=CD=\frac{BC}{2}$

In isosceles triangles when we draw a perpendicular to the third unequal side from the apex vertex then the third side is bisected by it or we can say that the third side is divided into two equal parts and the whole triangle is divided into two congruent triangles.