2014-I-09

$\begin{array}{ll}
\angle BAC = \angle CBD & \text{(given)} \\
\angle ACB = \angle BCD & \text{(common angle)} \\
\end{array}$

$\begin{array}{rll}
\angle ABC & = 180^\circ – \angle BAC – \angle ACB & \text{($\angle$ sum of $\Delta$)} \\
& = 180^\circ – \angle CBD – \angle BCD & \text{(proved)} \\
& = \angle BDC & \text{($\angle$ sum of $\Delta$)}
\end{array}$

Therefore, $\Delta ABC \sim \Delta BDC \text{ (A.A.A.)}$.

$\begin{array}{rcl}
\dfrac{AC}{BC} & = & \dfrac{BC}{DC} \\
\dfrac{25}{20} & = & \dfrac{20}{DC} \\
DC & = & 16 \text{ cm}
\end{array}$

Consider $\Delta BCD$,

$\begin{array}{rcl}
BD^2 + CD^2 & = & 12^2 + 16^2 \\
& = & 400
\end{array}$

and

$\begin{array}{rcl}
BC^2 & = & 20^2 \\
& = & 400
\end{array}$

Therefore, $BD^2 + CD^2 = BC^2$.

Hence by the converse of Pythagorus Theorem, $\Delta BCD$ is a right-angled triangle.