MHB Help Me Solve the Golden Ratio Question

[wrightarya]

Question is attached.
I don't know how to find the golden ratio

I really did try to solve this one, but i couldnt
please please help me out

 

[Prove It]

Question is attached.
I don't know how to find the golden ratio

I really did try to solve this one, but i couldnt
please please help me out

Call length AB = x and BC = y. Since AYXD is a square, that means length YB = x - y. Because the two rectangles are similar, their sides are in proportion, so

$\displaystyle \begin{align*} \frac{x}{y} &= \frac{y}{x - y} \\ x \left( x - y \right) &= y^2 \\ x^2 - x\,y &= y^2 \\ x^2 - x\,y + \left( -\frac{y}{2} \right) ^2 &= y^2 + \left( -\frac{y}{2} \right) ^2 \\ \left( x - \frac{y}{2} \right) ^2 &= \frac{4y^2}{4} + \frac{y^2}{4} \\ \left( x - \frac{y}{2} \right) ^2 &= \frac{ 5y^2}{4} \\ x - \frac{y}{2} &= \frac{\sqrt{5}\,y}{2} \\ x &= \frac{y + \sqrt{5}\,y}{2} \\ x &= \left( \frac{1 + \sqrt{5}}{2} \right) \, y \end{align*}$