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,

\[ \begin{aligned} {XZ \over ZW} &= {XY \over WY} \\ {XZ \over XY} &= {ZW \over WY} \\ \end{aligned} \]

\(\Box\)

# posted by rot13(Unafba Pune) @ 3:31 PM