Problem — Bisectors of the base angles are congruent

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Let $PQR$ be an isosceles triangle with base $QR$ and $PQ \cong PR$. The bisector of angle $\widehat{Q}$ meets side $PR$ at $S$, while the bisector of angle $\widehat{R}$ meets side $PQ$ at $T$. Prove that the segments $QS$ and $RT$ are congruent.

Solution

The triangle is isosceles with base $QR$, so the base angles are congruent: $\widehat{PQR} \cong \widehat{PRQ}.$ Since $QS$ and $RT$ are their bisectors, their halves are congruent as well: $\widehat{SQR} = \frac{1}{2}\,\widehat{PQR} \cong \frac{1}{2}\,\widehat{PRQ} = \widehat{TRQ}.$

Compare the triangles $QRS$ and $RQT$:

• $QR$ is a common side;
• $\widehat{SQR} \cong \widehat{TRQ}$ (halves of the base angles, just shown);
• $\widehat{SRQ} \cong \widehat{TQR}$ (these are the whole base angles $\widehat{PRQ}$ and $\widehat{PQR}$, which are congruent).

By the second congruence criterion (ASA) the two triangles are congruent: $\triangle QRS \cong \triangle RQT.$ From the corresponding sides (opposite the congruent angles $\widehat{SRQ}$ and $\widehat{TQR}$) it follows that $QS \cong RT. \qquad \blacksquare$