A 3:4:5 Triangle from a Square

Solution

Considering we will be using midpoints, I'm going to say the length of side AB is \(2n\).

\(BY^2=CY^2+BC^2=n^2+(2n)^2=5n^2\implies BY=\sqrt{5}n\)

Given that BX=BY (which can be proved in the same way) we can find angle XBA, which I will temporarily call \(\alpha\).

\(sin(\alpha)=\frac{AX}{XB}=\frac{n}{\sqrt{5}n}=\frac{1}{\sqrt{5}}\)

Now, let's use basic trigonometry on the triangle BYZ, with angle ZYB being \(2\alpha\).

\(YZ=BY cos(2\alpha)=\sqrt{5}n (1-2{sin}^2 (\alpha))=\sqrt{5}n (1-\frac{2}{5})=\frac{3\sqrt{5}n}{5}\)

That's two sides of the right-angled triangle, so use Pythagoras' Theorem to find BZ.

\(BZ^2=BY^2-ZY^2=(\sqrt{5}n)^2-({\frac{3\sqrt{5}n}{5}})^2=5n^2-\frac{9n^2}{5}=\frac{16n^2}{5}\)

Therefore \(BZ=\frac{4n}{\sqrt{5}}\) and the ratio YZ:ZB:BY is \(\frac{3n}{\sqrt{5}}:\sqrt{5}n:\frac{4n}{\sqrt{5}}\) which when multiplied through by \(\frac{\sqrt{5}}{n}\) gives \(3:5:4\)

The triangle is therefore a 3:4:5 triangle.