(1) Drawing the figure, I observe that ratios $\mathrm{AE}/\mathrm{AD}$ and $\mathrm{BC}/\mathrm{BE}$ are approximately equal, having a value of 1.6.

(2) From Pythagoras' Theorem I calculate $\mathrm{MC}$ (in the right-angled triangle $\mathrm{MBC}$):

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\mathrm{MC}}^2}& ={\mathrm{BC}}^2+{\mathrm{MB}}^2=1+1/4\\ \multicolumn{1}{c}{\mathrm{MC}}& =\sqrt{5}/2\end{array}.}$

So $\mathrm{AE}=(\sqrt{5}+1)/2$ and $\mathrm{BE}=(\sqrt{5}-1)/2$. The ratios are:

${\displaystyle \frac{\mathrm{AE}}{\mathrm{AD}}=\frac{\sqrt{5}+1}{2}}$

and

${\displaystyle \frac{\mathrm{BC}}{\mathrm{BE}}=\frac{1}{(\sqrt{5}-1)/2}=\frac{\sqrt{5}+1}{2}.}$

So, $\mathrm{AE}/\mathrm{AD}=\mathrm{BC}/\mathrm{BE}.$

(3) From this equality of ratios, I find out that

${\displaystyle \mathrm{BE}=\frac{\mathrm{AD}.\mathrm{BC}}{\mathrm{AE}}=\frac{1}{\varphi }}$

But $\mathrm{AE}=\mathrm{AB}+\mathrm{BE}$ so

${\displaystyle \varphi =1+\frac{1}{\varphi }.}$

(4) Substituting $\varphi =1$ the left hand side of this expression is less than the right hand side. If we increase the value given to $\varphi$ the left hand side increases and the right hand side decreases continuously. Substituting $\varphi =2$ the left hand side is greater than the right hand side so the value of $\varphi$ which satisfies this equation must lie between $1$ and $2$.

The two solutions of the equation can be found at the intersection of the cyan curve ( $y=1+1/x$) and magenta curve ( $y=x)$. Only the positive value is considered and it is approximately 1.618.

(5) Now, I solve the equation. It is equivalent to ${\varphi }^2-\varphi -1=0$ so the solutions are

${\displaystyle {\varphi }_1=\frac{1-\sqrt{5}}{2}}$

and

${\displaystyle {\varphi }_1=\frac{1+\sqrt{5}}{2}.}$

Only the second solution is valid because $\varphi >0$ .