For $\Delta ABC$ and $\Delta XYZ$,$\frac{AB}{XZ} = \frac{BC}{XY} = \frac{AC}{YZ}$. Then,the correspondence $ABC \leftrightarrow \ldots \ldots$ between them is a similarity.

$CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\Delta ABC$ and $\Delta EFG$ respectively. If $\Delta ABC \sim \Delta FEG,$ show that:
$(i) \frac{CD}{GH} = \frac{AC}{FG}$
$(ii) \Delta DCB \sim \Delta HGE$
$(iii) \Delta DCA \sim \Delta HGF$

It is given that $\triangle DEF \sim \triangle RPQ.$ Is it true to say that $\angle D = \angle R$ and $\angle F = \angle P?$ Why?

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YM}$ is an altitude to the hypotenuse $\overline{XZ}$. If $XM = 16ZM$,prove that $XY = 4YZ$.

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow QPR.$ If $m \angle P : m \angle Q : m \angle R = 2 : 3 : 4,$ then $m \angle A : m \angle B : m \angle C = \ldots$

$\Delta DEF$ and $\Delta PQR$ are similar triangles if $m \angle D = m \angle R$ and $\ldots \ldots$