In triangle ABC the medians AD,BE,CF intersect at G. Prove that ar(AGB) = 13 ar(ABC)
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Solution
A median bisects the triangle into two equal parts. Based on this, we can write, Area of triangle ABC = 2 ar ADB = 2 ar BEC = 2 ar ADC (all the following are areas) → ADB = BEC = ADC → AGB + GBD = AGC + GCD ; But, GBD = GCD (because D is midpoint of BC in triangle GCD) → AGB = AGC Similarly it can be proved that AGB = BGC. So, we have AGB = AGC = BGC. Again, ABC can also be written as AGB + AGC + BGC. → ABC = 3 AGB or AGB = 13 ABC. Proved.