Areas in a regular pentagon

We note that $EB$ divide $\triangle ABF$ into areas $[ABG]=a$ and $[BFG]=b$ , where $a$ is also the area of $\triangle BCF$ . We also note that $\triangle ABC$ and $\triangle BCF$ are similar. Let $BF=CF=1$ , then $AB=BC=2\cos 36^\circ$ and $\dfrac {[ABC]}{[BCF]} = \dfrac {2a+b}a = \blue{4 \cos^2 36^\circ}$ . Similarly, $\triangle ABF$ and $\triangle BFG$ are similar and $\dfrac {[ABF]}{[BFG]} = \dfrac {a+b}b = \blue{4 \cos^2 36^\circ}$ . Then we have

$\begin{aligned} \frac {2a+b}a & = \frac {a+b}b \\ 1 + \frac {a+b}a & = \frac ab + 1 \\ \implies \frac {a+b}a & = \frac ab = \varphi \end{aligned}$

Since $\dfrac {[ABF]}{[BCF]} = \dfrac {a+b}a = \dfrac ab = \boxed \varphi$ .

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Reference: golden ratio

It is known that the ratio of a diagonal of a regular pentagon to its side is $\varphi$ , hence, $AC=\varphi$ .

$\triangle ABF$ is isosceles , hence, $AF=AB=1.$ Consequently, $FC=AC-AF=\varphi -1=\frac{1}{\varphi }.$ $\ \$

The two coloured triangles share a common altitude from vertex $B$ . Therefore the ratio of their areas equals the ratio of the corresponding bases:

$\dfrac{\left[ ABF \right]}{\left[ BFC \right]}=\dfrac{AF}{FC}=\dfrac{1}{\frac{1}{\varphi }}=\boxed{\varphi}.$

Draw $BE$ and let the intersection of $AC$ and $BE$ be $G$ .

Let $CF = 1$ and $AF = x$ . By symmetry, $CF = BF = BG = AG = 1$ , and $AF = AB = BC = x$ .

Since $\triangle BCF \sim \triangle ACB$ , $\frac{BC}{CF} = \frac{AC}{AB}$ , or $\frac{x}{1} = \frac{x + 1}{x}$ , which solves to $x = φ$ .

Let the altitude of $\triangle ABF$ and $\triangle BCF$ from $B$ be $h$ . Then the ratio of the areas of those two triangles is $\frac{A_{\triangle ABF}}{A_{\triangle BCF}} = \frac{\frac{1}{2} \cdot AF \cdot h}{\frac{1}{2} \cdot CF \cdot h} = \frac{AF}{CF} = \frac{φ}{1} = \boxed{φ}$ .