Problem
GEO-B1-M04-P022 A Parallel Through the Intersection of Diagonals
In trapezoid \(ABCD\), bases \(AD\parallel BC\), \(AD=12\), \(BC=6\). The diagonals meet at \(O\). Through \(O\), a line parallel to the bases meets \(AB\) and \(CD\) at points \(X\) and \(Y\). Prove that \(OX=OY\), and find \(XY\).
Use \(DO:OB=AD:BC\), then compare triangles \(BOX\) and \(BDA\), and also \(DOY\) and \(DBC\).
Let \(AD=a\), \(BC=b\). From the similarity of triangles \(AOD\) and \(COB\), we get \(DO:OB=a:b\), so \(OB:DB=b:(a+b)\) and \(DO:DB=a:(a+b)\). In triangle \(BDA\), segment \(OX\parallel AD\), hence \(\frac{OX}{AD}=\frac{OB}{DB}=\frac{b}{a+b}\). Thus \(OX=\frac{ab}{a+b}\). In triangle \(DBC\), segment \(OY\parallel BC\), hence \(\frac{OY}{BC}=\frac{DO}{DB}=\frac{a}{a+b}\). Thus \(OY=\frac{ab}{a+b}\). Therefore \(OX=OY\). For \(a=12\), \(b=6\), we get \(OX=OY=\frac{12\cdot6}{18}=4\), so \(XY=8\).
A good 2–3 idea problem: trapezoid diagonals, similarity, then parallel sections.