In an isosceles right-angled triangle, the length of the hypotenuse is $24$. Find the area of the triangle.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 20$ and $BC = 21$,find $BM$. (in $.5$)

If $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow XZY$,then $BC^2 : YZ^2 = \ldots \ldots \ldots$ (Wait,the question asks for $ABC : XYZ = BC^2 : \ldots$ which refers to the ratio of areas). Given $\Delta ABC \sim \Delta XZY$,the ratio of their areas is equal to the ratio of the squares of their corresponding sides. Thus,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XZY)} = \frac{BC^2}{ZY^2}$.

In $\Delta ABC$,$A-N-B$,$A-M-C$ and $B-X-C$. $\overline{XM} \parallel \overline{AB}$ and $\overline{XN} \parallel \overline{AC}$. $\overline{MN}$ intersects $\overline{CB}$ at $T$. Prove that $TX^2 = TB \times TC$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $\frac{AM}{MB} = \frac{2}{3}$ and $AC = 25$,find $NC$.

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?