(2) By Pythagoras Theorem: ${\mathrm{MC}}^2=125$ so $\mathrm{MC}=5\sqrt{5}$, $\mathrm{AE}=5+5\sqrt{5}$ and $\mathrm{BE}=5\sqrt{5}-5$. Hence

${\displaystyle \frac{\mathrm{AE}}{\mathrm{AD}}=\frac{5+5\sqrt{5}}{10}=\frac{1+\sqrt{5}}{2}}$

and

${\displaystyle \frac{\mathrm{BC}}{\mathrm{BE}}=\frac{10}{5\sqrt{5}-5}=\frac{2}{\sqrt{5}-1}.}$

In order to write the last expression with a whole number in the denominator we multiply top and bottom by $\sqrt{5}+1$ which gives

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

Hence the two ratios are indeed equal.

(3) As the sides of the square are 1 unit, that is $\mathrm{BC}=1$ unit, and $\mathrm{BC}/\mathrm{BE}=\varphi$, then $\mathrm{BE}=1/\varphi$ and $\mathrm{AE}=1+1/\varphi$. We have also $\mathrm{AE}/\mathrm{AD}=\varphi$ so

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

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 $2$.

(4)Reading from the graph the values are approx 1.62 and -0.62

(5)Multiplying through by $\varphi$ this equation becomes:

${\displaystyle {\varphi }^2-\varphi -1=0}$

which has solutions $\varphi =(1\pm \sqrt{5})/2$ and taking the positive value this gives the golden ratio $\varphi =(1+\sqrt{5})/2$.