In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 2.5$,$MB = 3$ and $AN = 3.75$,find $AC$.

In $\Delta ABC$,$D$,$E$ and $F$ are the midpoints of $\overline{BC}$,$\overline{CA}$ and $\overline{AB}$ respectively. Then,$\text{Area}(BDEF) = \ldots \times \text{Area}(\Delta ABC)$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 12$ and $CM = 3$,find $BM$.

The ratio of the corresponding altitudes of two similar triangles is $\frac{3}{5}$. Is it correct to say that the ratio of their areas is $\frac{6}{5}$? Why?

In $\Delta XYZ$ and $\Delta MNO$,$\frac{XY}{MN} = \frac{XZ}{NO} = \frac{YZ}{MO}$. Then,the correspondence $\ldots$ is a similarity.

In $\Delta ABC$,$m \angle A = 90^\circ$. If $b = 12$ and $c = 35$,then $a = \ldots$