In $\triangle ABC$,$BC = AB$ and $\angle B = 80^{\circ}$. Then $\angle A$ is equal to: (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 the given figure,$XP = XS$,$XQ = XR$ and $\angle PXR = \angle SXQ$. Prove that $PQ = SR$.

In $\Delta PQR$,$X$ is the midpoint of $QR$. $XY$ and $XZ$ are altitudes from $X$ on $PQ$ and $PR$ respectively. If $XY = XZ$,then prove that $\Delta PQR$ is an isosceles triangle.

$ABC$ and $DBC$ are two triangles on the same base $BC$ such that $A$ and $D$ lie on the opposite sides of $BC$,$AB = AC$ and $DB = DC$. Show that $AD$ is the perpendicular bisector of $BC$.

In $\Delta ABC$,$AB = AC$ and $\angle CAD$ is an exterior angle of $\Delta ABC$ at vertex $A$. If ray $AP$ is the bisector of $\angle CAD$,then prove that $AP \parallel BC$.