Triangle $\mathrm{BXO}$ is similar to triangle $\mathrm{XPO}$ (angles ${30}^o,{60}^o,{90}^o$) so $\mathrm{BX}=\mathrm{BO}\cos 30=\frac{\sqrt{3}}{2}$ and $\mathrm{XO}=\mathrm{BO}\sin 30=\frac{1}{2}$, $\mathrm{OP}=\mathrm{XO}\sin 30=\frac{1}{4}$ and $\mathrm{XP}=\mathrm{PY}=\mathrm{XO}\cos 30=\frac{\sqrt{3}}{4}$ so $\mathrm{XY}=\frac{\sqrt{3}}{2}$.

Now consider triangle $\mathrm{OPZ}$. Angle $\mathrm{OPZ}={90}^o$. Therefore using Pythagoras' Theorem, $\mathrm{PZ}=\sqrt{(}1-\frac{1}{16})=\frac{\sqrt{1}5}{4}$. So $\mathrm{XZ}=\mathrm{XP}+\mathrm{PZ}=\frac{\sqrt{3}}{4}(1+\sqrt{5})$ and

${\displaystyle \frac{\mathrm{XZ}}{\mathrm{XY}}=\frac{1}{2}(1+\sqrt{5})}$

which is the Golden Ratio.




In the second diagram, taking the side of the square $\mathrm{XY}$ as 1 unit and using Pythagoras Theorem for triangle $\mathrm{OSY}$ gives $\mathrm{OS}=\frac{\sqrt{5}}{2}$ which is the radius of the circle. So $\mathrm{OZ}=\frac{\sqrt{5}}{2}$. Hence $\mathrm{XZ}=\frac{1}{2}(1+\sqrt{5})$ and

${\displaystyle \frac{\mathrm{XZ}}{\mathrm{XY}}=\frac{1}{2}(1+\sqrt{5})}$

again the Golden Ratio.




In the third diagram take the sides of the pentagon to be 1 unit long and the chords $\mathrm{XZ}$ and $\mathrm{WZ}$ etc. to be $x$ units.

Triangles $\mathrm{UXY}$, $\mathrm{YVZ}$ and $\mathrm{XZW}$ are similar isosceles triangles (angles of ${36}^o$, ${72}^o$ and ${72}^o$) with $\mathrm{VY}=\mathrm{VZ}=\mathrm{XY}=1$ and $\mathrm{YZ}=x-1$. We see that the ratio we want to find $\mathrm{XZ}/\mathrm{XY}=x$.

By similar triangles, taking the ratio of the long to the short sides, we get

${\displaystyle \frac{x}{1}=\frac{1}{x-1}}$

which gives the quadratic equation

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

with solutions $\frac{1}{2}(1+\sqrt{5})$ or $\frac{1}{2}(1-\sqrt{5})$. The latter is negative so does not make sense geometrically so the ratio must be the Golden Ratio:

${\displaystyle \frac{1}{2}(1+\sqrt{5}).}$