In $\Delta XYZ$,the bisector of $\angle Y$ intersects $\overline{XZ}$ at $M$. Then,which of the following holds true?

Given $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR.$ $\overline{AD}$ is a median in $\Delta ABC$ and $\overline{PM}$ is a median in $\Delta PQR.$ Prove that $\frac{AD}{PM} = \frac{AB}{PQ}.$

$A$ $8 \, m$ 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 $\Delta ABC$,$\overline{AD}$ is a median. If $AB = 8$ and $AD = BD = 8.5$,find $AC$.

$\Delta ABC \sim \Delta DEF$ for the correspondence $ABC \leftrightarrow FDE$. If $m \angle E + m \angle F = 130^{\circ}$,then $m \angle B = \dots$ (in $^{\circ}$)

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 15$ and $BM = 12.5$,find $BC$.