Given: A △ABC in which medians AD, BE and CF intersect at G. To prove: ar (△AGB) = ar (△BGC) = ar (△CGA) = 1/3 ar (△ABC) Proof: In △ABC, AD is the median. As a median of a triangle divides it into two triangles of equal area. ∴ ar ( △ABD) = ar ( △ACD) ....(i) In △GBC, GD is the median ∴ ar ( △GBD) = ar ( △GCD) .....(ii) Subtracting (ii) from (i), we get ar ( △ABD) - ar ( △GBD) = ar ( △ACD) - ar ( △GCD) ar ( △AGB) = ar ( △AGC) .....(iii) Similarly, ar ( △AGB) = ar ( △BGC) .....(iv) From (iii) and (iv), we get ar (△AGB) = ar (△BGC) = ar (△AGC) ...(v) But, ar (△AGB) + ar (△BGC) + ar (△AGC) = ar (△ABC) ....(vi) From (v) and (vi), we get 3 ar (△AGB) = ar (△ABC) ⇒ ar (△AGB) = 1/3 ar (△ABC) Hence, ar ( △ AGB) = ar ( △ AGC) = ar ( △ BGC) = 1/3 ar ( △ ABC) Yes, for the development of a society, education of each girl child is essential. An educated society always progresses.