3420.81 – Intersecting Medians

In $\triangle$ ABC, medians AD and BE intersect at G and ED is drawn. If the area of $\triangle$ EGD is k, find the area of $\triangle$ ABC.

Solution

Look at $\triangle$ BED. Medians intersect in the ratio 2:1, so set BG = 2x and GE = x. Thus the area of $\triangle$ BGD is 2 x area $\triangle$ GED = 2k.

Then look at $\triangle$ BEC. Area $\triangle$ BED = $\triangle$ CED.

$\triangle$ BED = 3k so $\triangle$ CED = 3k too.

Now consider $\triangle$ ADC. $\triangle$ AED = $\triangle$ CED, so $\triangle$ AEG = 2k.

Finally look at $\triangle$ BAC. $\triangle$ BAE = $\triangle$ BCE = 6k, so $\triangle$ BAG = 4k.

So we've got 12 k total.

(There might be snazzier ways to get it.)