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A pentagon ABCDE contains 5 triangles whose areas are each one. The triangles are ABC, BCD, CDE, DEA, and EAB. Find the area of ABCDE?

Is there a theorem for overlapping triangle areas?

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    Try to show that the pentagon has to be regular (at least I _think_ it has to be). Exploit that the triangles share bases and therefore heights. – 2017-02-01

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If $ABC$ and $BCD$ have the same area, then $BC\parallel AD$ and so on. It follows that $ABCDE$ is not necessarily a regular pentagon, but it is for sure the image of a regular pentagon through an affine map. Affine maps preserve the ratios between areas, hence it is enough to understand what is the area of a regular pentagon $ABCDE$ in which $[ABC]=1$. Such area is $\color{red}{\frac{5+\sqrt{5}}{2}}$. enter image description here $$\scriptstyle\text{A non-regular pentagon with } [ABC]=[BCD]=[CDE]=[DEA]=[EAB] $$

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    Is it possible to figure out the area of this pentagon algebraically? You can find the ratios between the triangles, but how can the ratios between triangles help you figure out the area? – 2017-02-19
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It was asked how to find the area algebraically.

Assume that the pentagon $ABCDE$ is regular and the area $[ABC] = 1$. Let $s$ be its side length. The angle at B is $3\pi/5$. The area of $ABC$ is then $s^2/2 \sin (3\pi/5)$, which must equal 1. The area of the pentagon is $1/4 \times 5 \times s^2 \times \cot (\pi/5)$ (see for example the Wikipedia article on regular polygons, https://en.wikipedia.org/wiki/Regular_polygon#Area). Therefore we have

$$ {[ABCDE] \over [ABC]} = {1/4 \times 5s^2 \times \cot (\pi/5) \over s^2/2 \times \sin(3\pi/5)} = {5 \over 2} {\cot (\pi/5) \over \sin(3\pi/5) } $$

So we need to evaluate $$x = {\cot (\pi/5) \over \sin(3\pi/5)}$$. Recalling the definition of the cotangent, we get

$$x = {\cos(\pi/5) \over \sin(\pi/5) \sin(2\pi/5)} $$

Now using the double angle formula, $\sin(2\pi/5) = 2 \sin(\pi/5) \cos(\pi/5)$ and so we have

$$x = {\cos \pi/5 \over 2 \sin^2 \pi/5 \cos \pi/5} = {1 \over 2 \sin^2 \pi/5} $$

and knowing that $$ \sin {\pi \over 5} = {1 \over 4} \sqrt{10 - 2\sqrt{5}}$$ (this is a standard fact one can look up) gives

$$ x= {8 \over (10 - 2 \sqrt{5})} = {8(10 + 2\sqrt{5}) \over (10+2\sqrt{5})(10-2\sqrt{5})} = {80 + 16 \sqrt{5} \over 80} = {5 + \sqrt{5} \over 5}$$

Plugging this back in we get, as desired,

$${[ABCDE] \over[ABC]} = {5 \over 2} {5 + \sqrt{5} \over 5} = {5 + \sqrt{5} \over 2}. $$