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    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.
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    Quote Originally Posted by kamaksh_ice View Post
    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:

    A=\frac{1}{2} \cdot base \cdot height. That means:

    A_{ABC}= \frac{1}{2} \cdot (AB) \cdot h_{1} and

    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

    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 \frac{1}{3} of the complete height.

    3. All these consideration are the same for all three triangles. Thus S must be the centroid.
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