Question

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

(Hint : In $\Delta APD$ and $\Delta BPC\overline{AD}=\overline{BC},\overline{AP}=\overline{BP}$ and $\angle PAD=\angle PBC=90°-60°=30°$]

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
$\angle PAD=\angle DAB-\angle PAB=90°-60°=30°$
$\angle PBC=\angle ABC-\angle PBA=90°-60°=30°$
$\therefore \angle DAP=\angle CBP$ ...(3)
So, from the S.A.S. congruency,
$\therefore \Delta APD\cong \Delta BPC$