$A$ $6.5 \, m$ long ladder leans on a wall to reach a height of $6 \, m$ on the wall. Then,the lower end of the ladder remains $\ldots \ldots \ldots \ldots \, m$ away from the wall.

Is it true to say that if in two triangles,an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle,then the triangles are similar? Give reasons for your answer.

$D$ is a point on side $QR$ of $\triangle PQR$ such that $PD \perp QR$. Will it be correct to say that $\triangle PQD \sim \triangle RPD$? Why?

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $m \angle A : m \angle C = 1 : 2$. If $BC = 4$,find $AC$.

If $\Delta XYZ \sim \Delta DEF$ for the correspondence $XYZ \leftrightarrow EFD$.
If $m \angle X : m \angle Y : m \angle Z = 2 : 3 : 5$,then in $\Delta DEF$,$\ldots \ldots$ is a right angle.

$\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$. $\overline{AD} \perp \overline{BC}$,$D \in \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$,$M \in \overline{YZ}$. Prove that $\frac{AD}{XM} = \frac{BC}{YZ}$.