2023-I-08

$\begin{array}{rcll}
\angle AEC & = & \angle BED & \text{(vert. opp. $\angle$)} \\
\angle ACE & = & \angle BDE & \text{(alt. $\angle$s, $AC$//$DB$)}
\end{array}$

$\therefore \Delta ACE \sim \angle BDE$ (A.A.)

$\begin{array}{rcll}
\dfrac{AC}{BD} & = & \dfrac{AE}{BE} & \text{(corr. sides, $\sim\Delta$s)} \\
\dfrac{10}{15} & = & \dfrac{20-BE}{BE} \\
2BE & = & 3(20-BE) \\
2BE & = & 60 -3BE \\
5BE & = & 60 \\
BE & = & 12\text{ cm}
\end{array}$

Furthermore,

$\begin{array}{rcll}
\dfrac{AC}{BD} & = & \dfrac{CE}{DE} & \text{(corr. sides, $\sim\Delta$s)} \\
\dfrac{10}{15} & = & \dfrac{7}{DE} \\
2DE & = & 21 \\
DE & = & 10.5\text{ cm}
\end{array}$

In $\Delta BDE$,

$\begin{array}{cl}
& DE^2 +BE^2 \\
= & (10.5)^2 +12^2\\
= & 254.25 \text{ cm}^2
\end{array}$

Also,

$\begin{array}{cl}
& BD^2 \\
= & 15^2 \\
= & 225\text{ cm}^2
\end{array}$

Since $DE^2 +BE^2 \neq BD^2$, then $\Delta BDE$ is not a right-angled triangle. (Converse of Pyth. Thm.)