Unit 1 Geometry Help

[KidwaiMaaz001]

Hi,
So I've figured out how to prove that AD/BD = BD/CD, but I don't get how to prove the second part of the question; that the aforementioned ratios are equal to φ - 1. How do I prove this?
Thanks.

[fun_jirachi]

Hey there!

If we have the radius of the circle being 1, and the side length of the square being \(s\), we can ascertain that \(s = \sqrt{1-\frac{s^2}{4}}\)  ie. we have that \(\frac{5s^2}{4} = 1 \implies s = \frac{2}{\sqrt{5}}\).
Hence, we have that \(AD = 1-\frac{1}{\sqrt{5}}\) and \(CD = 1+\frac{1}{\sqrt{5}}\), with \(BD = \frac{2}{\sqrt{5}}\). What we also note here is \(\phi - 1 = \frac{\sqrt{5}-1}{2}\) - we can also see that \(\frac{AD}{BD} = \frac{1-\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{\sqrt{5}-1}{2}\), and this holds true also for \(\frac{BD}{CD}\). Since \(\phi\) is a constant, we can just verify the ratios with the value given in the question - this holds true for any mathematical constant you might come across!

Hope this helps