In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YM}$ is an altitude to the hypotenuse $\overline{XZ}$. If $XM = \sqrt{12}$ and $ZM = \sqrt{3}$,find $YM$.

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

In $\Delta ABC$,$m\angle A + m\angle C = m\angle B$. If $AB = 7$ and $BC = 24$,then $AC = \ldots$

$A$ $8 \, m$ long bamboo tree standing erect on the ground breaks at the height of $3 \, m$ from the ground. The broken part of the tree remains attached to the trunk. Find the distance in $m$ between the top of the tree on the ground and the base of the tree.

In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect at $O$. $\overrightarrow{AO}$ intersects $\overline{BC}$ at $P$. Prove that $\frac{AB}{AC} = \frac{BP}{PC}$.

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