A Rectangle inside a Triangle!

Image from GeoGebra .

Let $x=\dfrac {|AD|}{|BD|}$

Let $\triangle ADE$ have area $y$ . Hence $[\triangle ABC] = (\dfrac {x+1}{x})^2y$ by property of similar triangles, since $DE \parallel BC$ .

Now note that the total shaded area is $\dfrac {25}{32} [\triangle ABC] = \dfrac {25}{32} (\dfrac {x+1}{x})^2y$ .

When the two shaded triangles at the bottom, $\triangle BDF$ and $\triangle CEK$ , are joined, the resultant triangle is similar to $\triangle ADE$ . Hence their total area is $\dfrac {y}{x^2}$ .

Equating, $y+\dfrac {y}{x^2} = \dfrac {25}{32} (\dfrac {x+1}{x})^2y$ $1+\dfrac {1}{x^2} = \dfrac {25}{32} \dfrac {x^2+2x+1}{x^2}$

Multiplying by $32x^2$ , $32x^2+32=25x^2+50x+25$ $7x^2-50x-7=0$ $(7x+1)(x-7)=0$

Since $x>0$ , we have $x=\boxed {7}$

Let $AM\bot BC$ . Let n= $\dfrac{AD}{DB}$ . p= $\dfrac{BC}{AB}$ .
The proportions in the similar triangles, because of DE || BC and DF || AM are as under.
$\dfrac{AD}{AB}= \dfrac n{n+1}=\dfrac{DE}{BC},~~~~Similarly ~~\dfrac{DF}{p}=\dfrac{AM}{p+p*n}.\\ \therefore ~ DE*DF=\{BC* \dfrac n{n+1}\} *\{AM*\dfrac p {p +p*n}\}= \dfrac n{(n+1)^2}*BC*AM \\ \implies~2* \Delta ABC= \dfrac{(n+1)^2} n* \boxed{~~~}DEKF ~~\implies~ \dfrac{(n+1)^2}{ 2n} =\dfrac {32}7(given)\\ Clearly~\dfrac {AD}{DB}=n=7$