Problem — A chord parallel to a side and the similar triangle

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In a triangle $PQR$ a point $M$ is taken on the side $PQ$, and from it the chord $MN$ is drawn parallel to the side $QR$, with $N$ on the side $PR$.

• (a) Prove that triangle $PMN$ is similar to triangle $PQR$.
• (b) Given that $\overline{PQ}=20$, $\overline{PR}=15$, $\overline{QR}=24$ and $\overline{PM}=8$, compute $\overline{PN}$ and $\overline{MN}$.

Solution

(a) Triangles $PMN$ and $PQR$ share the angle $\hat P$. Since $MN\parallel QR$, the angle $\widehat{PMN}\cong\widehat{PQR}$ (corresponding angles for the transversal $PQ$). Having two congruent angles, by the first similarity criterion the two triangles are similar: $PMN\sim PQR.$

(b) From the similarity the homologous sides are proportional with ratio $k=\frac{\overline{PM}}{\overline{PQ}}=\frac{8}{20}=\frac{2}{5}.$ Hence $\overline{PN}=k\cdot\overline{PR}=\frac{2}{5}\cdot 15=6,\qquad \overline{MN}=k\cdot\overline{QR}=\frac{2}{5}\cdot 24=\frac{48}{5}=9.6.$ $\boxed{\overline{PN}=6,\qquad \overline{MN}=9.6.}$