Ratio of areas of ABJ to BCDE

Question

We have two squares in a triangle like in the picture. We know that E divides AB into two halves and C divides FG into two halves. Can we somehow determine the ratio of ABJ to BCDE?

I have tried and tried again but there is just one variable that stands between me and the answer. The length of FG.

Answer 1

Clearly $\angle BAJ = 45^{\circ}$.

Let $BE=a$ and $FC=b$ and let $v$ be an altitude of $ABJ$ on $AB$.

• Then $DF = a-b$ and $FI = 2b$ so $a-b=2b\implies a=3b$.
• Since $ABJ\sim DGJ$ we have ${v-a\over v}= {a+b\over 2a}$ so $v=9b$.

Finnaly we have $${BCDE\over ABJ} = {a^2\over {2a\cdot v\over 2}} = {a\over v} = {3b\over 9b} = {1\over 3}$$

Answer 2

We have $AE=EB=ED$, so the angle $EAD\angle$ is $45^\circ$.
Consequently, we also have $DF=FI$, hence $DF=FG$, thus $F$ is midpoint and $C$ is quarter point of $DG$, yielding $$DG=\frac43DC=\frac23AB\\ FG=\frac13AB$$