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

$A$ $15 \, m$ long bamboo leans on a wall to reach the height of $12 \, m$ on the wall. Then,the lower end of the bamboo is $\ldots \ldots \, m$ away from the base of the wall.

In $\Delta ABC$,$m\angle B = 90^\circ$. If $AB = 12$ and $BC = 5$,then $AC = \dots$

In rectangle $ABCD$,$AC = 29$ and $AB + BC = 41$. Find $AB$ (given $AB > BC$).

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. Prove that $\frac{AB^2}{BC^2} = \frac{AM}{CM}$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$. $D$ is the midpoint of $\overline{BC}$ and $F$ is the midpoint of $\overline{AB}$. Prove that $AD^{2} + CF^{2} = \frac{5}{4} AC^{2}$.