# Ratio of areas of ABJ to BCDE

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.

• Observe that we know all angles, according to the conditions: $DE/AE=1$ and $BC/CG=3$. – Berci Nov 9 '19 at 11:31
• How do you know the length of CG though? – Pavol Komlos Nov 9 '19 at 11:36
• See my answer. ${{{{}}}}$ – Berci Nov 9 '19 at 11:40

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 $${BCDE\over ABJ} = {a^2\over {2a\cdot v\over 2}} = {a\over v} = {3b\over 9b} = {1\over 3}$$
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 $$DG=\frac43DC=\frac23AB\\ FG=\frac13AB$$