Given a triangle ABC and a point S inside.Show that if areas of triangle ABS, triangle BCS and triangle CAS are equal, then S is the centroid of ABC.
Hello,
1. The area of a triangle is calculated by:
$\displaystyle A=\frac{1}{2} \cdot base \cdot height$. That means:
$\displaystyle A_{ABC}= \frac{1}{2} \cdot (AB) \cdot h_{1}$ and
$\displaystyle A_{ABS}= \frac{1}{2} \cdot (AB) \cdot h_{2} = \frac{1}{3} \cdot \frac{1}{2} \cdot (AB) \cdot h_{1} = \frac{1}{2} \cdot (AB) \cdot \frac{1}{3} \cdot h_{1}$. Thus
$\displaystyle h_2 = \frac{1}{3} \cdot h_1$
2. A parallel to the side of a triangle through the centroid divides the height which is perpendicular to this side into 2 parts which have the ration 2 : 1. The part of the height which is connected with the side of the triangle is therefore $\displaystyle \frac{1}{3} $ of the complete height.
3. All these consideration are the same for all three triangles. Thus S must be the centroid.