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:


 A: This is a famous lemma from Archimedes.
                            
Let $O$ be the center of the circle and extend $C$ through $O$ to meet $E$ so that $\overline{CE}$ is a diameter. Join points as shown above.
Angles $\angle CBX$ and $\angle CEA$ are equal because they subtend the same arc $AC$ of the circle. 
Angle $\angle CAE$ subtends a diameter of the circle and is hence a right angle. 
Therefore triangles $\triangle BXC$ and $\triangle EAC$ are similar. This shows that $\angle ACE = \angle BCD$. 
Equal angles subtend equal arcs, and therefore arcs $BD$ and $AE$ are equal and hence segements $|BD|$ and $|AE|$ are equal.
From the Pythagorean theorem, we finish off as
$$\begin{align}(2r)^2 &= |AE|^2 + |AC|^2 
\\ &= |BD|^2 + (|AX|^2 + |CX|^2) 
\\ &= (|BX|^2 + |DX|^2) + (|AX|^2 + |CX|^2)\end{align}$$
Therefore
$$4r^2 = |AX|^2 + |BX|^2 + |CX|^2 + |DX|^2$$
as desired.
A: I will prove it for the isolated case of when the line $AB$ is the diameter. Cannot think how to generalize this now, but perhaps it will at least give some ideas.
Assume $AB$ is the diameter and $X$ is WLOG to the right of the origin, and the radius of the circle is $r > 0$. Let $\alpha$ denote the angle $XOC$ where $O$ is the center of the circle.
Then, $|AX| = r(1 + \cos \alpha)$ and $|XB| = r(1 - \cos \alpha)$, while $|CX| = |DX| = r \sin \alpha$. Note that
$|AX|^2 + |XB|^2 + |CX|^2 + |DX|^2
 = r^2 \left( (1+\cos\alpha)^2 +(1-\cos\alpha)^2 + 2\sin^2 \alpha\right)
 = r^2 \left( 2 + 2\cos^2 \alpha + 2\sin^2 \alpha \right)
 = 4r^2$
as desired.
A: Another way to solve it is introducing Cartesian system of coordinates with the origin at the circle center and axes along AB and CD. Let's say X has coordinates $(x, y)$ then $A(a, y), B(b, y), C(x, c)$ and $D(x, d)$. All four points lie on the circle, so the equation of the circle holds for them
$$A: a^2 + y^2 = r^2\\
B: b^2 + y^2 = r^2 \\
C: x^2 + c^2 = r^2 \\
D: x^2 + d^2 = r^2
$$
From $A$ and $B$ one can find that $a = -b$, and analogously $c = -d$. Then
$XA^2 + XB^2 + XC^2 + XD^2 = (a-x)^2 + (b-x)^2 + (c-y)^2 + (d-y)^2 = \\
a^2+b^2+c^2+d^2-2x(a+b)-2y(c+d)+2x^2+2y^2 = a^2+b^2+c^2+d^2+2x^2+2y^2$
which is sum of all LHS for the equations $A-D$ above and consequently $4r^2$
A: 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
$$4r^2 = a^2 + b^2 + c^2 + d^2$$
