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"Be a right triangle ABC with $\angle B=90º$. Two equilateral triangles $ABD$ and $BEC$ are drawn externally in the legs of the triangle $ABC$. Be $G,H$ and $F$ the midpoints of $BE$, $BC$ and $DC$. If the area of $ABC$ is $32$, then the area of $GHF$ is?"

I made the drawing with an arbitrary triangle $6,8,10$ in GeoGebra because i didn't know how to start in this problem, and i got that the area of $ABC$ is $4$ times the area of $GHF$ (the triangle $GHF$ is right too), so the answer will be $8$, but i want to know how to get this mathematically without trigonometry. Any hints?

enter image description here

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    $\begingroup$ Similarity? Because each side is halved if you form $\triangle PQR$ and thus the area will be $(\frac 12)^2=\frac 14$ times the original area. $\endgroup$
    – For the love of maths
    Oct 28, 2018 at 15:57
  • $\begingroup$ I know, but how can i get the similarity without graphing this in a program (Geogebra) $\endgroup$
    – SilentMath
    Oct 28, 2018 at 16:11
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    $\begingroup$ Look at $\overline{FH}$ as it relates to $\triangle CDB$, and at $\overline{GH}$ as it relates to $\triangle BEC$. Notice anything? $\endgroup$
    – Blue
    Oct 28, 2018 at 16:33
  • $\begingroup$ @Blue, oh, i see. I only need to determinate why the triangle $GHF$ is a right triangle. $\endgroup$
    – SilentMath
    Oct 28, 2018 at 17:27
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    $\begingroup$ @RodrigoPizarro: My hint helps with the right angle issue. Look at $\angle BHG$ and $\angle BCE$, as well as $\angle FHB$ and $\angle DBX$ (where $X$ is some point on the opposite side of $B$ from $C$). $\endgroup$
    – Blue
    Oct 28, 2018 at 17:34

2 Answers 2

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Since$$FH\parallel DB$$then$$\angle FHB=\angle DBJ=30^o$$And since$$HG\parallel CE$$then$$\angle BHG=\angle BCE=60^o$$enter image description hereTherefore$$\angle FHG=30^o+60^o=90^o$$And since$$FH=\frac12 DB=\frac12 AB$$and$$GH=\frac12 EC=\frac12 BC$$then in area$$\triangle FHG=\frac14 \triangle ABC=8$$

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(i) Verify that $\triangle(FGH)$ has a right angle at $H$. (ii) Determine the lengths of the legs of $\triangle(FGH)$.

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