In $\Delta ABC$,$m\angle B = 90^\circ$. If $AB = 12$ and $BC = 5$,then $AC = \dots$

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. Prove that $BD = \frac{BC \times AB}{AB + AC}$ and $DC = \frac{BC \times AC}{AB + AC}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $D$ is a point on $\overline{BC}$. Prove that $AD^{2} + BC^{2} = AC^{2} + BD^{2}$.

In $\Delta ABC$,$D$ and $E$ are the midpoints of $\overline{BC}$ and $\overline{AC}$ respectively. $\overline{AD}$ and $\overline{BE}$ intersect at $G$. Line $m$ passing through $D$ and parallel to $\overline{BE}$ intersects $\overline{AC}$ at $K$. Then,$AC = \ldots$ (in $EK$)

In $\Delta PQR, m\angle Q = 90^{\circ}$ and $\overline{QM}$ is a median. If $PQ = 20$ and $QR = 21$,find $QM$.

In $\Delta ABC$ and $\Delta PQR$,$m \angle A = m \angle R$ and $m \angle B = m \angle Q$. Then,the correspondence $ABC \leftrightarrow \ldots$ is a similarity.