Problem
GEO-B2-M07-P010 Midpoint from Equal Areas
#10
★★★☆☆ Level 3 of 5
Point \(P\) lies inside triangle \(ABC\), and line \(AP\) meets \(BC\) at \(D\). If \([PAB]=[PAC]\), prove that \(D\) is the midpoint of \(BC\).
Translate the equality of areas into the ratio \(BD:DC\).
We have \(\frac{BD}{DC}=\frac{[PAB]}{[PAC]}=1\). Therefore \(BD=DC\), so \(D\) is the midpoint of \(BC\).
A useful mini-lemma for proofs.