Angle bisector theorem via the sine rule

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In a triangle $ABC$, the bisector $BK$ of the angle at $B$ meets $AC$ at $K$. Using the sine rule, prove that $\frac{\overline{AK}}{\overline{AB}}=\frac{\overline{CK}}{\overline{CB}}.$

Solution

Let $\widehat{ABK}=\widehat{KBC}=\beta$, since $BK$ bisects $\widehat B$.

In triangle $ABK$, the sine rule gives: $\frac{\overline{AK}}{\sin\beta}=\frac{\overline{AB}}{\sin\widehat{AKB}}\quad\Longrightarrow\quad \frac{\overline{AK}}{\overline{AB}}=\frac{\sin\beta}{\sin\widehat{AKB}}.$

In triangle $CBK$, similarly: $\frac{\overline{CK}}{\overline{CB}}=\frac{\sin\beta}{\sin\widehat{BKC}}.$

But $\widehat{AKB}$ and $\widehat{BKC}$ are adjacent angles along segment $AC$, hence supplementary: $\widehat{AKB}+\widehat{BKC}=180°$, so $\sin\widehat{AKB}=\sin\widehat{BKC}$.

Comparing the two expressions gives: $\frac{\overline{AK}}{\overline{AB}}=\frac{\sin\beta}{\sin\widehat{AKB}}=\frac{\sin\beta}{\sin\widehat{BKC}}=\frac{\overline{CK}}{\overline{CB}}.$

$\boxed{\dfrac{\overline{AK}}{\overline{AB}}=\dfrac{\overline{CK}}{\overline{CB}}\quad\blacksquare}$