# Find the ratio of the area $EDPQ$ to the area of ABC

In triangle $$ABC$$,points $$E$$ and $$D$$ are on side $$AC$$ and point $$F$$ is on side $$BC$$ such that AE=ED=DC and $$BF:FC$$ =2:3. $$AF$$ intersects $$BD$$ and $$BE$$ at points $$P$$ and $$Q$$, respectively. Find the ratio of the area $$EDPQ$$ to the area of $$ABC$$

taken from the 2017 IMC held in India

I assumed ABD was equilateral so that the area of ABE would be equal to BED but didn’t gain anything from that

• Are you familiar with using mass points? – Anirudh Jul 6 at 18:43

Let $$k\in DC$$ such that $$FK||BD$$.
Thus, $$DK:KC=2:3.$$
Let $$DC=5x$$.
Thus, $$DK=2x$$ and $$AD=10x$$, which gives $$\frac{AP}{PF}=\frac{AD}{DK}=\frac{10x}{2x}=5.$$ Similarly, let $$M\in EC$$ such that $$FM||BE.$$
Thus, $$EM:MC=2:3,$$ which gives $$EM=4x$$ and $$\frac{AQ}{QF}=\frac{AE}{EM}=\frac{5x}{4x}=\frac{5}{4}.$$ From here we obtain: $$AQ:QP:PF=10:5:3.$$ Now, $$S_{\Delta BPQ}=\frac{5}{18}S_{\Delta ABF}=\frac{5}{18}\cdot\frac{2}{5}S_{\Delta ABC}=\frac{1}{9}S_{\Delta ABC}=\frac{1}{3}S_{\Delta BED},$$ which gives $$S_{EDPQ}=\frac{2}{3}S_{\Delta BED}=\frac{2}{9}S_{\Delta ABC}$$ and we are done!