Question
In the given figure, ABCD is a square and $\triangle PAB$ is an equilateral triangle.
(i) Prove that $\triangle APD \cong \triangle BPC$.
(ii) Show that $\angle DPC =15^{\circ}$

(i) Prove that $\triangle APD \cong \triangle BPC$.
(ii) Show that $\angle DPC =15^{\circ}$
