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In the figure, $ABCDEF$ is a regular hexagon, $FGH$ is an equilateral triangle, and $F$ is a common vertex of both figures. If the sides of both figures are the same, prove that :

$$\Delta AFH + \Delta EFG = \frac{HA · HG}{2}$$

(consider that the position of $\Delta FGH$ doesn't matter) enter image description here

What i've done -The area of $\Delta FGH$ is $\frac{ABCDEF}{6}$ -Notice how $HAEG$ is a cyclic cuadrilateral so we can use ptolemy's theorem, so $$ HE·GA = HA · GE + HG · AE$$

but i don't really know if it's the correct way to apply it.

Is it possible to do this exercise without any trigonometrical functions?

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    The shaded area is $HG・AF$. – 2017-02-16
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    How do you know that? – 2017-02-16
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    Since $AFH=EFG=$ right triangle. What are sides of both figures? – 2017-02-16
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    @TakahiroWaki Why $AFH=EFG$? – 2017-02-16
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    "both figures" is $HF=AF$? – 2017-02-16

1 Answers 1

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In the picture $P\neq A$ is a point on the line $AF$ such that $|AF|=|PF|$. Also, by hypothesis $|AF|=|HF|$, which implies that $F$ is the circumcenter of $\triangle AHP$, so it is straight at $H$.

enter image description here

Now, observe that $$\measuredangle PFH=\measuredangle PFG+\measuredangle GFH=\measuredangle PFG+\frac{\pi}3=\measuredangle PFG+\measuredangle EFP=\measuredangle EFG$$ Now, from the given hypothesis we have $|GF|=|FH|$ and $|EF|=|AF|$ and $|AF|=|PF|$ by construction, then $\triangle GEF$ and $\triangle HPF$ are equal.

I will use the notation $[\mathcal{A}]$ to denote the area of the figure $\mathcal{A}$, so $$\text{Shaded area }=[AFH]+[EFG]=[AFH]+[HPF]=[AHP]=\frac{|HA|\cdot|HP|}{2}$$ And from $|HP|=|GE|$ we get $$\text{Shaded area }=\frac{|HA|\cdot|GE|}{2}$$