# Right triangle with 2 equilateral triangles

"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?

• 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. – Mohammad Zuhair Khan Oct 28 '18 at 15:57
• I know, but how can i get the similarity without graphing this in a program (Geogebra) – Rodrigo Pizarro Oct 28 '18 at 16:11
• Look at $\overline{FH}$ as it relates to $\triangle CDB$, and at $\overline{GH}$ as it relates to $\triangle BEC$. Notice anything? – Blue Oct 28 '18 at 16:33
• @Blue, oh, i see. I only need to determinate why the triangle $GHF$ is a right triangle. – Rodrigo Pizarro Oct 28 '18 at 17:27
• @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 $C$). – Blue Oct 28 '18 at 17:34

Since$$FH\parallel DB$$then$$\angle FHB=\angle DBJ=30^o$$And since$$HG\parallel CE$$then$$\angle BHG=\angle BCE=60^o$$Therefore$$\angle FHG=30^o+60^o=90^o$$And since$$FH=\frac12 DB=\frac12 AB$$and$$GH=\frac12 EC=\frac12 BC$$then in area$$\triangle FHG=\frac14 \triangle ABC=8$$
(i) Verify that $$\triangle(FGH)$$ has a right angle at $$H$$. (ii) Determine the lengths of the legs of $$\triangle(FGH)$$.