A triangle and a square in a regular pentagon

(0) Labeling angles, you'll get

(1) $\angle AEB=\angle BAC=36^o=\frac{\pi}{5}^r$ by the fact that $\Delta AEB$ is an isosceles triangle

and

(2) $\angle EAF=\angle EFA=72^o=\frac{2\pi}{5}^r$ $(\angle EAF = \angle EAB- \angle BAC)$ , making $\Delta AEF$ an isosceles triangle.

(3) After that, draw $AP$ such that $AP$ is perpendicular to $BE$ and since earlier we get $\angle AEB=36^o=\frac{\pi}{5}^r$ , by trigonometry, we get that $EP$ is $\frac{1+\sqrt5}{4}x$ . (Let the pentagon's side length equal to $x$ .

(4) By using the fact $AP$ is perpendicular to $BE$ and $AEB$ is an isosceles triangle, $BE=2AP=\frac{1+\sqrt5}{2}$ .

(5) Because $\Delta AEF$ is an isosceles triangle, $AE=EF=x$ .**

(6) Connect $CE$ , this will split $CDEF$ into two identical triangles, $\Delta EFC$ and $\Delta CDE$

(7) After labeling more angles, $\Delta EFC$ is similar to \(\Delta AFB).

**(8)** From (4) and (5), we can now find \(FB\) which is $EB-EF= \frac{1+\sqrt5}{2}x-x=\frac{\sqrt5-1}{2}x$ .

(9) If two triangles are similar, the ratio of the area is equal to the ratio of the adjacent side lengths squared, from (7), $\Delta EFC$ is similar to $\Delta AFB$ and (8), the ratio of the adjacent side lengths (In this case, $\frac{FB}{EF}$ ) is $\frac{\frac{\sqrt5-1}{2}x}{x}$ = $\frac{\sqrt5-1}{2}$ , thus it follows that $\frac{[\Delta AFB]}{[\Delta CEF}=\frac{\sqrt5-1}{2}^2=\frac{3-\sqrt5}{2}$ .

(10) From (6), $\frac{[\Delta EFC]}{[CDEF]}$ = $\frac{1}{2}$ and italicized claim , it then follows $\frac{[\Delta AFB]}{[\Delta CDEF]}=\frac{[\Delta AFB]}{2[\Delta CEF]}=\frac{3-\sqrt5}{4}$

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Bonus from the challenge: This method uses $\cos36$ only once!