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

$CD$ 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:
$(i) \frac{CD}{GH} = \frac{AC}{FG}$
$(ii) \Delta DCB \sim \Delta HGE$
$(iii) \Delta DCA \sim \Delta HGF$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then the correspondence $ADB \leftrightarrow \ldots$ between $\Delta ADB$ and $\Delta BDC$ is a similarity.

If $\Delta ABC \sim \Delta MNP$ for the correspondence $ABC \leftrightarrow MNP,$ the perimeter of $\Delta ABC$ is $18$ and the perimeter of $\Delta MNP$ is $27.$ If $AB = 8,$ then $MN = \dots$

In $\Delta ABC$, $P$ and $Q$ are the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. If the area of $\Delta APQ = 12 \sqrt{3}$, then the area of $\Delta ABC = \ldots$ (in $\sqrt{3}$)

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?