Problem — A chord through the centroid parallel to a side

Text

Let $G$ be the centroid of a triangle $ABC$. Through $G$ the line parallel to the side $BC$ is drawn, meeting $AB$ at $X$ and $AC$ at $Y$.

• (a) Prove that the chord $XY$ equals $\tfrac{2}{3}$ of the side $BC$.
• (b) Given that $\overline{BC}=15$, compute $\overline{XY}$.

Solution

(a) Let $M$ be the midpoint of $BC$; the median $AM$ passes through the centroid $G$, which divides it so that $\frac{\overline{AG}}{\overline{AM}}=\frac{2}{3}.$ Since $XY\parallel BC$, triangles $AXY$ and $ABC$ share the angle $\hat A$ and have $\widehat{AXY}\cong\widehat{ABC}$ (corresponding angles): by the first criterion $AXY\sim ABC$. In the similarity the ratio of homologous sides equals the ratio of the respective altitudes from $A$; but the line $XY$ is at the same distance from $A$ as $G$, while $BC$ is at the same distance as $M$, so the ratio is $k=\frac{\overline{AG}}{\overline{AM}}=\frac{2}{3}.$ Consequently $\overline{XY}=k\cdot\overline{BC}=\frac{2}{3}\,\overline{BC}.\qquad\blacksquare$

(b) With $\overline{BC}=15$: $\overline{XY}=\frac{2}{3}\cdot 15=10.$ $\boxed{\overline{XY}=10.}$