2014-I-09

$\begin{array}{ll}
\angle BAC = \angle CBD & \text{(已知)} \\
\angle ACB = \angle BCD & \text{(公共角)} \\
\end{array}$

$\begin{array}{rll}
\angle ABC & = 180^\circ – \angle BAC – \angle ACB & \text{($\Delta$的內角和)} \\
& = 180^\circ – \angle CBD – \angle BCD & \text{(已證)} \\
& = \angle BDC & \text{($\Delta$ 的內角和)}
\end{array}$

所以，$\Delta ABC \sim \Delta BDC \text{ (A.A.A.)}$。

$\begin{array}{rcl}
\dfrac{AC}{BC} & = & \dfrac{BC}{DC} \\
\dfrac{25}{20} & = & \dfrac{20}{DC} \\
DC & = & 16 \text{ cm}
\end{array}$

考慮 $\Delta BCD$，

$\begin{array}{rcl}
BD^2 + CD^2 & = & 12^2 + 16^2 \\
& = & 400
\end{array}$

以及

$\begin{array}{rcl}
BC^2 & = & 20^2 \\
& = & 400
\end{array}$

所以，$BD^2 + CD^2 = BC^2$。

根據畢氏定理的逆定理，$\Delta BCD$ 為一直角三角形。