Geometry Prove - two perpendicular lines in a circle In a circle of radius r, two lines (AB and CD) are perpendicular to each other and meet at X. 
Show that:



Let $O$ be the center of the circle, and let $M$ and $N$ be the midpoints of segments $AB$ and $CD$, respectively. Write $a:=|XA|$, $b:=|XB|$, $c:=|XC|$, $d:=|XD|$, $m := |OM|$, $n:=|ON|$.
Then (noting that $OMXN$ is a rectangle),
$$\begin{align}
|AM| = \frac{1}{2}(a+b) &\qquad |ON|=|MX| = \frac{1}{2}|a-b| \\[5pt]
|CN| = \frac{1}{2}(c+d) &\qquad |OM|=|NX| = \frac{1}{2}|c-d| 
\end{align}$$
Now, $\triangle AMO$ and $\triangle CNO$ are right triangles with hypotenuses $|AO| = |CO| = r$, so that
$$\begin{align}
r^2 &= |AM|^2 + |OM|^2 = \frac{1}{4}(a+b)^2 + \frac{1}{4}(c-d)^2 \\[5pt]
r^2 &= |CN|^2 + |ON|^2 = \frac{1}{4}(c+d)^2 + \frac{1}{4}(a-b)^2
\end{align}$$
Adding these equations causes the "$ab$" and "$cd$" terms to cancel; multiplying through by $2$ gives
$$4r^2 = a^2 + b^2 + c^2 + d^2$$
