In $\Delta ABC$,$m \angle A = 90^{\circ}$,$AB = 8$ and $AC = 15$. Then,$BC = \ldots$.

In $\Delta ABC$,the correspondences $ABC \leftrightarrow BAC$ and $ABC \leftrightarrow ACB$ are similarities. Then,$\Delta ABC$ is a/an $\ldots \ldots \ldots \ldots$ triangle.

Prove that the area of the semicircle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

In $\Delta ABC$,$\overline{AD}$,$\overline{BE}$ and $\overline{CF}$ are medians. Prove that $3(AB^2 + BC^2 + AC^2) = 4(AD^2 + BE^2 + CF^2)$.

In the figure,if $\angle A = \angle C$,$AB = 6 \, cm$,$BP = 15 \, cm$,$AP = 12 \, cm$,and $CP = 4 \, cm$,then find the lengths of $PD$ and $CD$ (in $cm$).

In $\Delta ABC$,$m \angle B = 90$ and $D$ is the midpoint of $\overline{AC}$. Then,$BD = \dots$