In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 12$ and $BM = 12$,find $AC$.

Which of the following correctly matches the information in Part $I$ and Part $II$?

Part $I$	Part $II$
$1.$ In $\Delta ABC$,$\angle B$ is a right angle and $\overline{BM}$ is a median.	$a. AB^2 + BC^2 = 2(BD^2 + CD^2)$
$2.$ In $\Delta ABC$,$\angle A$ is a right angle and $\overline{AD}$ is an altitude.	$b. BC = \frac{1}{2} AB$
$3.$ In $\Delta ABC$,$m\angle C = 90^\circ$ and $m\angle A = 30^\circ$.	$c. AC^2 = CD \cdot BC$
$4.$ In $\Delta ABC$,$\overline{BD}$ is a median.	$d. BM = \frac{1}{2} AC$

In $\square ABCD$,$AB = AD$. The bisector of $\angle BAC$ intersects $\overline{BC}$ at $E$ and the bisector of $\angle DAC$ intersects $\overline{DC}$ at $F$. Prove that $\overline{EF} \parallel \overline{BD}$.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB = 12$,$AC = 8$ and $BD = 9$,find $DC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AB = \sqrt{10}$ and $AM = 2.5$,then $MC = \ldots$

It is given that $\triangle DEF \sim \triangle RPQ.$ Is it true to say that $\angle D = \angle R$ and $\angle F = \angle P?$ Why?