In parallelogram $ABCD$,$AB^{2} + AD^{2} = 200$ and $BD = 12$. Then $AC = \ldots$

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $\overline{PR}$. Then,$\angle PQD \cong \ldots \ldots \ldots$

In $\Delta ABC$,$AB > AC$ and $D$ is the midpoint of $\overline{BC}$. $\overline{AM} \perp \overline{BC}$ and $M \in \overline{BC}$. Prove that $AB^{2} - AC^{2} = 2 \cdot BC \cdot DM$.

In $\Delta XYZ$,the midpoints of the sides are $A, B$ and $C$. In $\Delta ABC$,the midpoints of the sides are $P, Q$ and $R$. If the area of $\Delta ABC$ is $24$,find the area of $\Delta XYZ$ and $\Delta PQR$.

Foot of a $10\,m$ long ladder leaning against a vertical wall is $6\,m$ away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches. (in $m$)

The correspondence $ABC \leftrightarrow QPR$ between $\Delta ABC$ and $\Delta PQR$ is a similarity. If $m \angle A : m \angle B : m \angle C = 15 : 8 : 7$,find the measures of all the angles of $\Delta PQR$.