Question

If the medians of a AABC intersect at G, then show that ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = 1/3 ar(ΔABC). Thinking Process Use the property that median of a triangle divides it into two triangles of equal area. Further, apply above property by considering different triangles and prove the required result.

Answer 1

PROOF:

i dont have the triangle sign so i will use <ABC> like this to indicate that i mean triangle

for denoting area i will write like this [ABC]

let D be the point where the median from A intersects on BC

in <ABC>

[ABD]=[ACD]

as the height is same and the base is same

(1/2*b*h)

[GBD]=[GCD]

as the height and base is same

subract these 2 equations and you get

[ABD]-[GBD]=[ACD]-[GCD]

Hence

[AGB]=[AGC] (refer to a diagram for better understanding)

similarly, we can prove that

[BGC]=[AGC]

[CGB]=[AGB]

hence we can say that <ABC> is divided into 3 equal parts and hence

[BGC]=1/3 * [ABC]

hence proved