Problem
GEO-B1-M06-P026 A Point Inside a Parallelogram
#26
★★★★☆ Level 4 of 5
Point \(P\) lies inside parallelogram \(ABCD\). Prove that \(S_{PAB}+S_{PCD}=\frac{1}{2}S_{ABCD}\).
Sides \(AB\) and \(CD\) are equal and parallel; the sum of the heights from \(P\) to them equals the height of the parallelogram.
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 \(AB\). Therefore \(S_{PAB}+S_{PCD}=\frac{1}{2}AB\cdot h_1+\frac{1}{2}CD\cdot h_2\). Since \(AB=CD\), this is \(\frac{1}{2}AB(h_1+h_2)=\frac{1}{2}S_{ABCD}\).
A strong and useful idea: the sum of distances to two parallel sides is constant.