Golden Ratio & Fibonacci - Charles de Gaulle 13-unit two-beamed cross problem.

Question

Here is the question:

The two-beamed cross, made popular by Charles de Gaulle, is formed from 13 unit squares as shown below. A straight line $BC$ drawn through point $A$ divides the cross in such a way that the total areas on each side of the line are equal. Show that $$\frac{PB}{BQ} = \phi$$

[where $\phi = 1.618\dots$ is the golden ratio].

Cheers, Axiom.

Answer 1

Let $BQ=x$ and $CF=y$. Then the upper part has area equals to $$6.5=5+area(\triangle ABQ)+area(\triangle ACD)=5+\frac x2+\frac{2+y}2.$$ Thus we have $x+y=1$, hence $PB=y$.

On the other hand $\triangle ABE$ and $\triangle ACD$ are similar, so $$AE:AD = BE:CD,$$ or $$x = \frac{1}{2+y}=\frac{1}{3-x}.$$ From there, one should be able to compute $x$ and verify that $\dfrac{1-x}{x}=\phi$.