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
Best Answer
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!