Problem
GEO-B1-M06-P021 Three Equal Areas
#21
★★★★☆ Level 4 of 5
Point \(P\) lies inside triangle \(ABC\). It is known that \(S_{PAB}=S_{PBC}=S_{PCA}\). Prove that \(P\) is the intersection point of the medians of the triangle.
Use equality of two areas with a common base to obtain a median.
From \(S_{PAB}=S_{PCA}\), line \(AP\) passes through the midpoint of \(BC\), so \(AP\) is a median. From \(S_{PAB}=S_{PBC}\), line \(BP\) passes through the midpoint of \(AC\), so \(BP\) is a median. Therefore \(P\) is the intersection of two medians, hence the centroid of the triangle.
Key strong idea: equal areas can determine the centroid.