5 triangles with the same area inside a pentagon 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?

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