Problem
GEO-B1-M08-P017 Point Inside a Parallelogram
#17
★★★★☆ Level 4 of 5
Point \(P\) lies inside parallelogram \(ABCD\). Prove that \(S_{PAB}+S_{PCD}=\frac{1}{2}S_{ABCD}\).
Add the distances from \(P\) to two parallel sides.
Let the distances from \(P\) to lines \(AB\) and \(CD\) be \(h_1\) and \(h_2\). Then \(h_1+h_2\) equals the height of the parallelogram to base \(AB\). Since \(AB=CD\), we have \(S_{PAB}+S_{PCD}=\frac{1}{2}ABh_1+\frac{1}{2}ABh_2=\frac{1}{2}S_{ABCD}\).
Requires seeing areas, although the figure is a parallelogram.