2018-I-13

$\begin{array}{rcll}
\angle ABE + \angle ECD & = & 180^\circ & \text{(alt. $\angle$s, $AB//DC$)} \\
90^\circ + \angle ECD & = & 180^\circ & \text{(given)} \\
\angle ECD & = & 90^\circ \\
\angle ABE & = & \angle ECD
\end{array}$

Also,

$\begin{array}{rcll}
\angle BAE + \angle ABE & = & \angle AEC & \text{(ext. $\angle$ of $\Delta$)} \\
\angle BAE + \angle ABE & = & \angle AED + \angle DEC \\
\angle BAE + 90^\circ & = & 90^\circ + \angle DEC & \text{(given)}\\
\angle BAE & = & \angle DEC
\end{array}$

Hence, we have

$\begin{array}{rcll}
\angle AEB & = & 180^\circ – \angle ABE – \angle BAE & \text{($\angle$ sum of $\Delta$)} \\
& = & 180^\circ – \angle ECD – \angle CDE & \text{(proved)} \\
& = & \angle EDC & \text{($\angle$ sum of $\Delta$)}
\end{array}$

$\therefore \Delta ABE \sim \Delta ECD \ \text{(A.A.A.)}$.

1. Since $\Delta ABE \sim \Delta ECD$, we have

   $\begin{array}{rcll}
   \dfrac{AB}{EC} & = & \dfrac{AE}{ED} & \text{(corr. sides, $\sim \Delta$)} \\
   \dfrac{15}{36} & = & \dfrac{25}{ED} \\
   ED & = & 60 \text{ cm}
   \end{array}$

   Since $\angle ECD = 90^\circ$ (proved in (a)), we have

   $\begin{array}{rcll}
   CD^2 & = & ED^2 – EC^2 & \text{(Pyth. Thm.)} \\
   CD^2 & = & 60^2 – 36^2 \\
   CD & = & 48\text{ cm}
   \end{array}$

2. It is given that $\angle AED = 90^\circ$, then the area of $\Delta ADE$

   $\begin{array}{cl}
   = & \dfrac{1}{2} \times AE \times ED \\
   = & \dfrac{1}{2} \times 25 \times 60 \\
   = & 750\text{ cm}^2
   \end{array}$

3. It is given that $\angle AED = 90^\circ$, then we have

   $\begin{array}{rcll}
   AD^2 & = & AE^2 + ED^2 & \text{(Pyth. Thm.)} \\
   AD^2 & = & 25^2 + 60^2 \\
   AD & = & 65\text{ cm}
   \end{array}$

   Let $h\text{ cm}$ be the height with respect to the base $AD$. Note that $h\text{ cm}$ is the shortest distance from $E$ to $AD$. By the result of (b)(ii), we have

   $\begin{array}{rcl}
   \dfrac{1}{2} \times h \times AD & = & 750 \\
   \dfrac{1}{2} \times h \times 65 & = & 750 \\
   h & = & 23.076\ 923\ 08 \\
   h & > & 23
   \end{array}$

   Since the shortest distance from $E$ to $AD$ is larger that $23\text{ cm}$, then there is no point $F$ lying on $AD$ such that th distance between $E$ and $F$ is less than $23\text{ cm}$.