Problem
GEO-B1-M06-P017 A Parallel Line and Area
#17
★★★☆☆ Level 3 of 5
In triangle \(ABC\), through point \(D\) on side \(BC\), a line parallel to \(AC\) meets \(AB\) at point \(E\). It is known that \(BD:DC=2:3\). Find \(S_{BDE}:S_{ABC}\).
Triangles \(BDE\) and \(BCA\) are similar.
Since \(DE\parallel AC\), \(\triangle BDE\sim\triangle BCA\). We have \(BD:BC=2:5\). Therefore areas are in the ratio of squares of corresponding sides: \(S_{BDE}:S_{ABC}=4:25\).
Reinforces the relation between similarity and areas in another orientation.