Problem
GEO-B1-M08-P020 A Parallel and the Remaining Area
#20
★★★★☆ 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 \(E\). Prove that \(S_{BDE}:S_{ADEC}=4:21\).
First find the ratio \(S_{BDE}:S_{ABC}\).
Since \(DE\parallel AC\), \(\triangle BDE\sim\triangle BCA\). The similarity ratio is \(BD:BC=2:5\), so \(S_{BDE}:S_{ABC}=4:25\). The remaining part \(ADEC\) has area \(21\) parts out of \(25\). Hence \(S_{BDE}:S_{ADEC}=4:21\).
Mixed method: similarity gives the small area, then area chasing is needed.