Problem
GEO-B1-M06-P018 A Point on a Median
#18
★★★☆☆ Level 3 of 5
In triangle \(ABC\), median \(AM\) is drawn to side \(BC\). Point \(P\) lies on \(AM\). Prove that \(S_{PAB}=S_{PAC}\).
Line \(AP\) passes through the midpoint of \(BC\).
Since \(P\) lies on \(AM\), line \(AP\) is the median and passes through the midpoint \(M\) of side \(BC\). Triangles \(PAB\) and \(PAC\) have common base \(AP\). Points \(B\) and \(C\) are at equal distances from line \(AP\), so the areas are equal.
A needed idea for proving centroid facts by areas.