Ans: (a) $\Delta AED\sim\Delta BEC$, $AE=6\text{ cm}$ (b) Yes
- $\Delta ADE \sim \Delta BCE$. Hence, we have
$\begin{array}{rcl}
\dfrac{AE}{BE} & = & \dfrac{DE}{CE} \\
\dfrac{AE}{8} & = & \dfrac{15}{20} \\
AE & = & 6\mbox{ cm}
\end{array}$ - In $\Delta ABE$,
$\begin{array}{rcl}
AE^2 + BE^2 & = & 6^2 + 8^2 \\
& = & 100 \mbox{ cm}^2
\end{array}$$\begin{array}{rcl}
AB^2 & = & 10^2 \\
& = & 100 \mbox{ cm}^2
\end{array}$Since $AE^2+BE^2=AB^2=100\mbox{ cm}^2$, by the converse of Pythagoras Theorem, $\Delta ABE$ is a right-angled triangle with $\angle AEB=90^\circ$.
Hence, $AC\perp BD$.
