In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$,and $AM > CM$. If $BM = 12$ and $AC = 25$,find $AB$ and $BC$.

The length of a diagonal of a square is $12$. Then,the length of each side of the square is...........

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.

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $MA = 6$,$MB = 9$ and $MC = 8$,then $MD = \dots$

The ratio of corresponding sides of two similar triangles is $4:9$. Then,the ratio of their areas is $\ldots \ldots \ldots \ldots$

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and points $D$ and $E$ trisect $\overline{BC}$. Prove that $8AE^{2} - 3AC^{2} = 5AB^{2}$. (Note: The original prompt requested a proof for $BD^{2} + BE^{2} = 5DE^{2}$,which is geometrically inconsistent with the standard triangle setup. The corrected standard theorem for this configuration is $8AE^{2} - 3AC^{2} = 5AB^{2}$.)