Geometry - Pentagon This is a tough program I have hard time to find the answer. Can anyone help me? Thank you very much in advance.
Problem - In regular pentagon $ABCDE$, point $M$ is the midpoint of side $AE$, and segments $AC$ and $BM$ intersect at point $Z$. If $ZA = 3$, what is the value of $AB$? Express your answer in simplest radical form.
This is a drawing for the problem:

 A: Choose the coordinate system so that $A = (0,0)$, $B = (\rho,0)$ and the other three vertices lie in the upper half plane. Identify the plane with complex numbers, the
five sides of the pentagon corresponds to
$$( AB, BC, CD, DE, EA ) \quad\leftrightarrow\quad (\rho,\rho\omega,\rho\omega^2,\rho\omega^3,\rho\omega^4)$$
where $\omega = e^{2\pi i/5}$ is the primitive $5^{th}$ root of unity.
Under this identification,
$$B = AB = \rho,\; C = AB+BC = \rho(1 + \omega)
\quad\text{ and }\quad
M = -\frac12 EA = -\frac{\rho}{2} \omega^4 = -\frac{\rho}{2} \bar{\omega}$$
Since $Z$ lies on the intersection of $AC$ and $BM$, there exists real numbers $\lambda, \mu$ such that
$$AZ = \lambda AC = (1-\mu) AB + \mu AM$$
This impies
$$\frac{Z}{\rho} = \lambda( 1 + \omega) = (1-\mu) - \mu\frac{\bar{\omega}}{2}
= \left[1 - \mu \left( 1 + \frac{\omega + \bar{\omega}}{2} \right)\right] + \frac{\mu}{2}\omega
$$
The square bracket in RHS is clearly a real number. By comparing the real
and imaginary part of the last equality, we get:
$$\lambda = 1 - \mu\left(1 + \frac{\omega + \bar{\omega}}{2} \right) = \frac{\mu}{2}
\implies \lambda = \frac{1}{3 + \omega + \bar{\omega}}
$$
This implies
$$|AB| = \rho
= \left|\frac{AZ}{\lambda(1 + \omega)}\right| 
= 3 \left|\frac{3+\omega+\bar{\omega}}{1+\omega}\right|
= 3\left(\frac{3 + 2\cos\frac{2\pi}{5}}{2\cos\frac{\pi}{5}}\right)
= 3\left(\frac{2+\phi}{\phi}\right) = 3\sqrt{5}
$$
A: Let $F$ be the intersection of $AC$ and $BE$. By angle chasing, we obtain the following picture:

Now $\triangle BAF$ and $\triangle BEA$ are similar, so 
$$BF:BA = BA:BE,$$
or 
$$\frac yx = \frac{x}{x+y}=\frac{1}{1+y/x}.\tag{1}$$
Solving that for $y$ in terms of $x$ we have 
$$\frac yx = \frac{\sqrt{5}-1}2.$$
Now apply Menelaus' Theorem to $\triangle AEF$ with $M$, $Z$, $B$ colinear we get
$$\frac{EM}{MA}\cdot \frac{ZA}{ZF}\cdot \frac{BF}{BE} = 1.\tag{2}$$
Note that $EM = MA$,
$$\frac{BF}{BE} = \frac{y}{x+y}=1-\frac{x}{x+y}=\frac{3-\sqrt{5}}2,$$
and
$$\frac{ZA}{ZF} = \frac{3}{y-3}.$$
Thus, (2) becomes
$$\frac{y-3}{3} = \frac{3-\sqrt{5}}2.$$
So
$$ y = \frac{9-3\sqrt{5}}{2}+3 = \frac{15-3\sqrt{5}}2.$$
And we get 
$$ x = \frac xy \cdot y = \frac{\sqrt{5}+1}2 \cdot \frac{15-3\sqrt{5}}2$$

Note: 


*

*I'm sure there are more elegant solutions.

*One can avoid Menelaus' Theorem by connecting $M$ with the midpoint $N$ of $AF$.

