A "simple" proof about triangles and relativity (similar triangles and Pythagoras)

Question

In "ABC of relativity" Bertrand Russel claims that the following is easy to prove but I'm stuck.

In the image $OD=OC$ and $OY=OX$, he claims that $OC^2-OX^2=SZ^2-RQ^2$

Also every angle that seems of 90 degrees is indeed a 90 degree angle

Answer 1

First we note that $YS$, $DR$, $XQ$ and $CZ$ are parallel, and all the right angled triangles are similar. Because $OC=OD$ and $OX=OY$, we can also say that $$ OS =XQ\quad\text{and}\quad OR=CZ\qquad(1)$$

Then we start in earnest: \begin{align*} SZ^2-RQ^2 &= (OZ-OS)^2-(OQ-OR)^2,\\ &= (OZ^2-OR^2) + (OS^2-OQ^2)+2OZ.OS -2OQ.OR\\ &= (OZ^2-CZ^2) + (XQ^2-OQ^2)+2OZ.XQ -2OQ.CZ\\ &= OC^2-OX^2 + 2(OZ.XQ -OQ.CZ), \end{align*} where to get to the third line we have used identities (1) and to get to the final line we applied the Pythagorean theorem on the bracketed terms in the preceding line.

The final bracketed term is zero thanks to triangle similarity: $$ \frac{OQ}{XQ} = \frac{OZ}{CZ}.$$