2016-I-13

$\begin{array}{rcll}
\angle ADC & = & \angle AEB & \text{(given)} \\
AD & = & AE & \text{(sides opp. eq. $\angle$s)}
\end{array}$

In $\Delta ACD$ and $\Delta ABE$,

$\begin{array}{rcll}
\angle ADC & = & \angle AEB & \text{(given)} \\
AD & = & AE & \text{(proved)} \\
CD & = & CE + ED & \\
& = & BD + ED & \text{(given)} \\
& = & BE &
\end{array}$

Therefore $\Delta ACD \cong \Delta ABE$ (S.A.S.).

1. Note that $AD = AE$. Then $\Delta ADE$ is an isosceles triangle. Note also that $DM = EM$. Then $AM$ is the median of the isosceles $\Delta ADE$. With the base $DE$, the median is the same as the height of the isosceles $\Delta ADE$. Hence, $\angle AMD = 90^\circ$. Then we have

   $\begin{array}{rcll}
   AM^2 & = & AD^2 – DM^2 & \text{(Pyth. Thm.)} \\
   AM & = & \sqrt{15^2 – (\dfrac{18}{2})^2} & \\
   AM & = & 12 \text{ cm}
   \end{array}$

2. Consider $\Delta ABM$,

   $\begin{array}{rcll}
   AB^2 & = & AM^2 + BM^2 & \text{(Pyth. Thm.)} \\
   AB & = & \sqrt{12^2 + (9+7)^2} & \\
   AB & = & 20 \text{ cm} &
   \end{array}$

   Note that $AD = AE = 15\text{ cm}$. Consider $\Delta ABE$,

   $\begin{array}{rcl}
   AB^2 + AE^2 & = & 20^2 + 15^2 \\
   & = & 625
   \end{array}$

   $\begin{array}{rcl}
   BE^2 & = & (7+18)^2 \\
   & = & 625
   \end{array}$

   $\because AB^2 + AE^2 = BE^2 = 625$,

   $\therefore$ by the converse of Pythagoras Theorem, $\Delta ABE$ is a right-angled triangle.