The ratio of the corresponding sides of two similar triangles is $4:9$. Then,the ratio of their areas is $\ldots \ldots$.

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

In trapezium $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. Prove that the correspondence $MAB \leftrightarrow MCD$ between $\Delta MAB$ and $\Delta MCD$ is a similarity.

$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

In $\Delta ABC$,$\Delta PQR$,and $\Delta XYZ$,the correspondences $ABC \leftrightarrow QPR$ and $PQR \leftrightarrow YZX$ are similarities. Then,the correspondence $ABC \leftrightarrow \dots$ is a similarity.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 6$,then $BM = \dots$