In △ABC, medians AD & BE intersect at G and ED is drawn. If the area of △EGD = k, find the area of △ABC.
Look at △BED. Medians intersect in the ratio 2:1, so set
Thus the area of △BGD is two times the area
Then look at △BEC. Area
Now consider △ADC.
Finally, look at △BAC.
So we’ve got 12k total.
(There might be snazzier ways to get to it.)