Problem
GEO-B1-M06-P025 Small Area With a Parallel Line
#25
★★★★☆ Level 4 of 5
In triangle \(ABC\), point \(D\) lies on \(BC\), with \(BD:DC=2:3\). Through \(D\), a line parallel to \(AC\) meets \(AB\) at point \(E\). Prove that \(S_{BDE}:S_{ADEC}=4:21\), where \(ADEC\) is the remaining part of the triangle.
First find \(S_{BDE}:S_{ABC}\).
Triangles \(BDE\) and \(BCA\) are similar because \(DE\parallel AC\). The similarity ratio is \(BD:BC=2:5\). Therefore \(S_{BDE}:S_{ABC}=4:25\). The remaining part \(ADEC\) has area \(21\) parts out of \(25\). Hence \(S_{BDE}:S_{ADEC}=4:21\).
The task requires not stopping at the ratio to the whole, but passing to the remaining part.