2009-I-11

$\begin{array}{ll}
\angle ADE & \\
= 180^\circ – \angle ACD – \angle CAD & \text{($\angle$ sum of $\Delta$)} \\
= 180^\circ – \angle ACD – \angle BCE & \text{(given)} \\
= \angle ACB & \text{(adj. $\angle$s on a st. line)}
\end{array}$

In $\Delta ABC$ and $\Delta AED$,

$\begin{array}{ll}
AC = AD & \text{(given)} \\
BC = ED & \text{(given)} \\
\angle ACB = \angle ADE & \text{(proved)}
\end{array}$

$\therefore \Delta ABC \cong \Delta AED$ (SAS).

1. In $\Delta ABF$ and $\Delta DEA$,

   $\begin{array}{ll}
   \angle ABF = \angle DEA & \text{(corr. $\angle$s, $\cong \Delta$s)} \\
   \angle AFB = \angle DAE & \text{(alt. $\angle$s, $AD//BC$)} \\
   \angle FAB & \\
   = 180^\circ – \angle ABF – \angle AFB & \text{($\angle$ sum of $\Delta$)} \\
   = 180^\circ – \angle DEA – \angle DAE & \text{(proved)} \\
   = \angle ADE & \text{($\angle$ sum of $\Delta$)}
   \end{array}$

   $\therefore \Delta ABF \sim \Delta DEA$ (AAA).

2. $\Delta CBA$ and $\Delta CEF$.