Geometry, Triangles In the figure, $BC$ is parallel to $DE$. If area of ∆ $PDE$ is $3/7$ of area of ∆ $ADE$, then what is the ratio of $BC$ and $DE$? I tried finding ratios of height of ∆ $ABC$, $PDE$ & $BPC$, and tr...--prophetes.ai

Geometry, Triangles In the figure, $BC$ is parallel to $DE$. If area of ∆ $PDE$ is $3/7$ of area of ∆ $ADE$, then what is the ratio of $BC$ and $DE$? I tried finding ratios of height of ∆ $ABC$, $PDE$ & $BPC$, and trying to figure out some commonality, but it didn't work out. P.s. it is not my homework. Ratio is 5:2. Not sure how. ![enter image description here](

We may assume $$A=(0,0),\quad B=(1,0),\quad C=(0,1), \quad D=(r,0),\quad E=(0,r)$$ for some $r\in\>]0,1[\>$. Intersecting $EB$ with $C D$ gives $P=\bigl({r\over1+r},{r\over1+r}\bigr)$. $ED$ and $PA$ intersect orthogonally at the midpoint $M=\bigl({r\over2},{r\over2}\bigr)$ of $ED$. The ratio of the two triangle areas in question is therefore given by $${|PM|\over |MA|}={\sqrt{2}\bigl({r\over 1+r}-{r\over2}\bigr)\over\sqrt{2}\,{r\over2}}={1-r\over1+r}\ .$$ Since this ratio has to be ${3\over7}$ it follows that $r={2\over5}$.