2021-I-08

$\begin{array}{rcll}
\angle CAE & = & \angle BDE & \text{(given)} \\
\angle AEC & = & \angle DEB & \text{(common $\angle$s)} \\
\end{array}$

$\therefore \Delta ACE \sim \Delta DBE$ (AA).

1. $\begin{array}{rcl}
   AC^2 + AE^2 & = & 25^2 + 60^2 \\
   & = & 4225
   \end{array}$

   $\begin{array}{rcl}
   CE^2 & = & 65^2 \\
   & = & 4225
   \end{array}$

   $\because AC^2+AE^2=CE^2$,

   $\therefore \Delta ACE$ is a right-angled triangle. (converse of Pyth Theorem).

2. Since $\Delta ACE \sim \Delta DBE$ and $\Delta ACE$ is a right-angled triangle, then $\Delta DBE$ is also a right-angled triangle.

   $\begin{array}{rcll}
   \dfrac{DE}{AE} & = & \dfrac{BD}{CA} & \text{(corr. sides, $\sim\Delta$s)} \\
   \dfrac{DE}{60} & = & \dfrac{15}{25} \\
   DE & = & 36 \text{ cm}
   \end{array}$

   $\therefore$ the area of $\Delta BDE$

   $\begin{array}{cl}
   = & \dfrac{1}{2} \times BD \times DE \\
   = & \dfrac{1}{2} \times 15 \times 36 \\
   = & 270\text{ cm}^2
   \end{array}$