In rhombus $ABCD$,$AC = 24$ and $BD = 70$. Find the perimeter of rhombus $ABCD$.

In the figure,if $LM \parallel CB$ and $LN \parallel CD$,prove that $\frac{AM}{AB} = \frac{AN}{AD}$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM : AB = 2 : 3$ and $AC = 15$,then $NC = \ldots$

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 $8AE^{2} - 3AC^{2} = 5AB^{2}$.)

In $\Delta ABC,$ the bisectors of $\angle B$ and $\angle C$ intersect $\overline{AC}$ and $\overline{AB}$ at $D$ and $E$ respectively. If $\overline{DE} \parallel \overline{BC},$ prove that $\Delta ABC$ is an isosceles triangle.

In $\Delta XYZ, m\angle Y = 90^{\circ}$. If $XY = YZ$ and $XZ = 10$,then $XY = \ldots$