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 *x**x*

Thus the area of △BGD is two times the area *k*

Then look at △BEC. Area *k**k*

Now consider △ADC.

*k*

Finally, look at △BAC.

*k**k*

So we’ve got 12*k* total.

(There might be snazzier ways to get to it.)