In $\triangle ABC$,$AB = AC$ and $\angle B = 50^{\circ}$. Then $\angle C$ is equal to (in $^{\circ}$)

"If two sides and an angle of one triangle are equal to two sides and an angle of another triangle,then the two triangles must be congruent." Is the statement true? Why?

Point $M$ lies on the perpendicular bisector of $PQ$. Also $M$ does not lie on $PQ$. Prove that $M$ is equidistant from $P$ and $Q$.

In $\Delta XYZ$,$\angle Y = 90^{\circ}$ and $XY = YZ$,then $\angle X = \dots$ (in $^{\circ}$)

In $\Delta ABC$,if $\angle A > \angle B$,then $AC \ldots \ldots BC$.

In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect at $P$. $A$ line drawn through $P$ and parallel to $BC$ intersects $AB$ at $X$ and $AC$ at $Y$. Prove that $XY = XB + YC$.