Problem
GEO-B1-M03-P020 Parallels to Two Medians
In triangle \(ABC\), point \(P\) lies on side \(AC\). Through \(P\), draw lines parallel to the medians from vertices \(A\) and \(C\). They meet sides \(BC\) and \(AB\) at \(E\) and \(F\), respectively. Prove that the medians from \(A\) and \(C\) divide segment \(EF\) into three equal parts.
Draw midlines through side midpoints and use parallel segments inside triangles. It is also convenient to use coordinates along side \(AC\).
Let \(K\) and \(L\) be the midpoints of sides \(BC\) and \(AB\); the medians are \(AK\) and \(CL\). It is enough to prove the statement in a convenient affine model, because parallelism and ratios on one line are preserved. Take coordinates \(A=(0,0)\), \(C=(1,0)\), \(B=(0,1)\). Then \(K=\left(\frac{1}{2},\frac{1}{2}\right)\), \(L=\left(0,\frac{1}{2}\right)\). Let \(P=(t,0)\).
The line through \(P\) parallel to \(AK\) has direction \((1,1)\) and meets \(BC\) at \(E=\left(\frac{1+t}{2},\frac{1-t}{2}\right)\). The line through \(P\) parallel to \(CL\) has direction \((-1,\frac{1}{2})\) and meets \(AB\) at \(F=\left(0,\frac{t}{2}\right)\). Parametrize \(EF\) as \(Z(\lambda)=E+\lambda(F-E)\).
Median \(AK\) has equation \(y=x\). Substituting \(Z(\lambda)\) gives \(\lambda=\frac{1}{3}\). Median \(CL\) has equation \(y=\frac{1-x}{2}\). Substitution gives \(\lambda=\frac{2}{3}\). Thus the intersection points of the medians with \(EF\) divide the segment at \(\frac{1}{3}\) and \(\frac{2}{3}\), so the three parts are equal.
This is a star-style problem within the introductory module. It can be reserved for a strong group or moved to a ladder.