In $\Delta PQR$,$PQ = PR$ and $\angle R = 40^{\circ}$,then $\angle P = \dots$ (in $^{\circ}$)

$CDE$ is an equilateral triangle formed on side $CD$ of a square $ABCD$ (see figure). Show that $\triangle ADE \cong \triangle BCE$.

In $\Delta ABC$,$AB = AC$ and $AD$ is an altitude to the base $BC$. Prove that $D$ is the midpoint of $BC$.

In a triangle $ABC$,$D$ is the mid-point of side $AC$ such that $BD = \frac{1}{2} AC$. Show that $\angle ABC$ is a right angle.

In $\Delta ABC$, $\angle B = \angle C$, $AB = 5 \text{ cm}$, and $BC = 8 \text{ cm}$. Then, the perimeter of $\Delta ABC$ is: (in $\text{ cm}$)

$O$ is a point in the interior of a square $ABCD$ such that $\triangle OAB$ is an equilateral triangle. Show that $\triangle OCD$ is an isosceles triangle.