Golden Rectangle where width is equal to 2 units

Question

Question reads: Commencing with a square ABCD of side length 2 units, locate E, the midpoint of DC. With centre E and radius EB draw an arc to intersect DC produced at Y. At Y construct YX perpendicular to DY to meet AB produced at X.

a) Show that rectangle AXYD is Golden, and, b) Show that rectangle BXYC is also Golden

I've been working on this question on and off for what feels like a few hours and I can't get beyond the first question. I've attempted to use the following proportion:

YC:CD = YD:YC

to yield proof that the length of the rectangle follows the golden ratio, but I've found myself stuck.

I'm at the point where I can say that the length of the rectangle is equal to (2+2√5)/2, or twice that of Phi. Is that sufficient proof to deem that sides AX and DY follow the golden ratio, as they are equal to the width of the rectangle * Phi?

And for question 2, would I be using a similar strategy to prove that the rectangle BXYC is also Golden? If not, where would I begin?

Cheers.

Answer 1

Yes, the length of the rectangle is $AX=\frac{2+2\sqrt{5}}{2}=1+\sqrt{5}$.

So $\frac{AX}{AD}=\frac{1+\sqrt{5}}{2}$, the golden ratio.

For Question 2, as $BX=\sqrt{5}-1$, we just have to show that $\frac{BC}{BX}=\frac{2}{\sqrt{5}-1}$ is equal to $\phi$, and it is not difficult.