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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?

enter image description here

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

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  • $\begingroup$ Observe that we know all angles, according to the conditions: $DE/AE=1$ and $BC/CG=3$. $\endgroup$ – Berci Nov 9 '19 at 11:31
  • $\begingroup$ How do you know the length of CG though? $\endgroup$ – Pavol Komlos Nov 9 '19 at 11:36
  • $\begingroup$ See my answer. ${{{{}}}}$ $\endgroup$ – Berci Nov 9 '19 at 11:40
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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 $${BCDE\over ABJ} = {a^2\over {2a\cdot v\over 2}} = {a\over v} = {3b\over 9b} = {1\over 3}$$

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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 $$DG=\frac43DC=\frac23AB\\ FG=\frac13AB$$

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