I was experimenting with regular polygons and I've found an interesting property with the regular heptagon in particular. Here it is:
Let $ABCDEFG$ be a regular heptagon. Now extend rays $GA$ and $CB$ to met at a point $H.$ It seems that the sum of the areas of $FED, GCBA, ABH$ are equal to $FDCG.$ Is this true?
There doesn't seem much to do aside from directly computing the areas. If you put the heptagon on the coordinate plane with $E=(1,0), F=(\cos(\frac{2\pi}{7}), \sin(\frac{2\pi}{7})),$ and so on, some computation leads to the area of $FDCG$ is $8\sin(\frac{\pi}{14})\cos^3(\frac{\pi}{14})\cos(\frac{\pi}{7}).$ The area of $FED$ is $(\sin(\frac{3\pi}{14})-1)(-\cos(\frac{3\pi}{14}))$, the area of $GCBA$ is $(\cos(\frac{\pi}{7})-\sin(\frac{\pi}{14}))(\sin(\frac{\pi}{7})+\cos(\frac{\pi}{14}))$, The area of $ABH$ is $\sin(\frac{\pi}{7})(1+2\sin(\frac{\pi}{14})-\cos(\frac{\pi}{7})).$ It seems really messy to check that the sum of the latter three values is equal to the former, how should I approach doing this?
The triangles $CHG$ and $CEG$ are congruent because they are both isosceles, share a side and a bit of angle chasing reveals that their base angles are equal. Let $[\ldots]$ denote the area of a polygon. For simplicity, set $[DEF] = x$. Note that then also $[CDE] = x$ and $[EFG] = x$. You conjecture that $$ [CEG] + x = [CDFG]. $$ To this end, by decomposing $CDEFG$ with the lines $CE$ and $EG$, we find $$ [CDEFG] = [CEG] + 2x. $$ Decomposing $CDEFG$ with the line $FD$, we obtain $$ [CDEFG] = [CDFG] + x. $$ Subtracting these two equations proves your result.