Question

In the adjacent figure $ABCD$ is a square and $△APB$ is an equilateral triangle. Prove that $△APD\cong △BPC$

Answer 1

Given that $APB$ is an equilateral triangle, so the sides of an equilateral triangle are equal.

$AP=BP$ ...... (1)

and $ABCD$ is a square. All the sides of a square are equal.

$AD=BC$ .....(2)

and

$\angle PAD=\angle DAB-\angle PAB={90}^0-{60}^0={30}^0$

$\angle PBC=\angle ABC-\angle PBA={90}^0-{60}^0={30}^0$

$\therefore \angle DAP=\angle BPC$ ......(3)

So, from the $S.A.S.$ congruency,

$\therefore \Delta APD\cong \Delta BPC$