Problem
GEO-B2-M07-P019 A Parallel Line and Areas
#19
★★★★☆ Level 4 of 5
In triangle \(ABC\), through point \(P\in AC\), a line parallel to \(BC\) is drawn, meeting \(AB\) at \(Q\). Prove that \(\frac{[APQ]}{[ABC]}=\left(\frac{AP}{AC}\right)^2\).
Triangles \(APQ\) and \(ABC\) are similar.
Since \(PQ\parallel BC\), triangles \(APQ\) and \(ABC\) are similar with ratio \(\frac{AP}{AC}\). Areas of similar triangles are in the ratio of the squares of similarity ratios, hence \(\frac{[APQ]}{[ABC]}=\left(\frac{AP}{AC}\right)^2\).
Connects areas with similarity through an area ratio.