$ABCD$ is a rectangle. If $AB = 7.5$ and $AC = 19.5$,find $BC$.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. Prove that $BD = \frac{BC \times AB}{AB + AC}$ and $DC = \frac{BC \times AC}{AB + AC}$.

$\square ABCD$ is a trapezium in which $\overline{ AD } \| \overline{ BC }, \overline{ AC } \cap \overline{ BD }=\{ P \} .$ If $PD =9, PA =5$ and $PB =7.2,$ then $AC =\ldots \ldots \ldots \ldots$

In the given figure,$\angle BAC = 90^{\circ}$ and $AD \perp BC$. Then,

$\Delta ABC \sim \Delta PQR$ 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 $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $BD = 2.5$,$AB = 5$ and $AC = 4.2$,find $DC$.