Problem — Statement and proof of the angle-bisector theorem

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State and prove the internal angle-bisector theorem: in a triangle the bisector of an interior angle divides the opposite side into two segments proportional to the other two sides of the triangle.

Solution

Statement. In triangle $RST$ the bisector of the angle $\hat R$ meets the side $ST$ at the point $U$. Then $\frac{\overline{SU}}{\overline{UT}}=\frac{\overline{RS}}{\overline{RT}}.$

Proof. From the vertex $T$ draw the line parallel to the bisector $RU$; it meets the extension of the side $SR$ at a point $V$. Consider the transversals $SV$ and $ST$ cut by the two parallels $RU$ and $TV$: by the Thales theorem \frac{\overline{SU}}{\overline{UT}}=\frac{\overline{SR}}{\overline{RV}}. \tag{1} It remains to show that $\overline{RV}=\overline{RT}$, that is, that triangle $RTV$ is isosceles on the base $TV$. Since $RU\parallel TV$:

• $\widehat{SRU}\cong\widehat{RVT}$ (corresponding angles),
• $\widehat{URT}\cong\widehat{RTV}$ (alternate interior angles).

But $RU$ is the bisector of $\hat R$, so $\widehat{SRU}\cong\widehat{URT}$; hence $\widehat{RVT}\cong\widehat{RTV}$. Triangle $RTV$ has two congruent base angles, therefore it is isosceles and $\overline{RV}=\overline{RT}$. Substituting into $(1)$: $\frac{\overline{SU}}{\overline{UT}}=\frac{\overline{RS}}{\overline{RT}}.\qquad\blacksquare$