Given any triangle with an angle bisector, like:
It can be shown that:
| |
\[
\begin{aligned}
{XZ \over XY} = {ZW \over WY} \\
\end{aligned}
\] |
Why ? By the
Law of Sines,
| |
\[
\begin{aligned}
{XZ \over \sin{\left(180-(\alpha + \beta)\right)}} &= {ZW \over \sin{\alpha}} \\
{WY \over \sin{\alpha}} &= {XY \over \sin{(\alpha + \beta)}} \\
\end{aligned}
\] |
But \(\sin{\left(180-(\alpha + \beta)\right)} = \sin{(\alpha + \beta)}\). So,
| |
\[
\begin{aligned}
{XZ \over \sin{\left(180-(\alpha + \beta)\right)}} &= {XZ \over \sin{(\alpha + \beta)}} = {ZW \over \sin{\alpha}} \\
\therefore {XZ \over ZW} &= {\sin{(\alpha + \beta)} \over \sin{\alpha}} \\
\end{aligned}
\] |
Recall
| |
\[
\begin{aligned}
{WY \over \sin{\alpha}} &= {XY \over \sin{(\alpha + \beta)}} \\
\therefore {\sin{(\alpha + \beta)} \over \sin{\alpha}} &= {XY \over WY} \\
\end{aligned}
\] |
Therefore,
| |
\[
\begin{aligned}
{XZ \over ZW} &= {XY \over WY} \\
{XZ \over XY} &= {ZW \over WY} \\
\end{aligned}
\] |
\(\Box\)
