Why are the diameters I've made perpendicular? In the following picture, point A was found by drawing a line from point C through point B.  As it happens, point A is ninety degrees from point E.  Can someone explain why this is true?  
A gif of the process used to make this image can be found here.
 A: I've added some additional lines in my copy of your diagram below. I won't provide a formal proof but just point out some geometric properties which you may find useful. I'm switching the question but it remains equivalent to your original question. Without line ABC, the construction is possible with the diameter of the largest circle parallel to the chord from point $C$. Due to symmetry, $DE$ will be perpendicular to $AD$. The question therefore is, will a line through $CB$ pass through $A$? This is equivalent to your question.
The sketch below can be used to prove this via the congruency of triangles $FDG$ and $HDG$, and angles subtended by the same arcs ($GB$ and $AF$) at the center $H$ and $D (2x)$ being double the angle at the circumference $C (x)$. What this does is show the line AC is colinear with line BC, both being at angle x to the chord FC, hence the extension of CB passes through A.
It then follows that point A will determine the diameter of the larger circle through AD which will be perpendicular to DE.

A: I've renamed points for no good reason ...

A little angle chasing gives that $\angle OQP \cong \angle OTP$ and $\angle POQ\cong\angle ROQ$. Also, in $\bigcirc Q$,
(central) $\angle PQU$ and (inscribed) $\angle PTU$ subtend $\stackrel{\frown}{PU}$, so that $\angle UTP = \frac12\angle OQP = \frac12\angle OTP$.
Thus, $\overline{OU}$ and $\overline{TU}$ are bisectors of respective angles in $\triangle OTP$, so that $U$ is the incenter of that triangle. This implies that $\overline{PV}$ bisects $\angle OPS$. Finally $\triangle OPV$ is isosceles,
$\overline{OVP} \cong \angle OPV \cong \angle VPS$, whence $\overline{OV}\parallel\overline{PS}$, and the result follows. $\square$
A: According to the link, the construction seems to be as follows: On the circumference of a given circle, taking points $F$, $G$ at random, draw a circle with center $F$ and radius $FG$ meeting the given circle again at $H$, and with center $H$ and radius $HF$ draw another circle meeting the given circle again at $C$. Join $GC$, intersecting the two equal circles at $K$, $L$, and join $FK$ and $HL$, extended to meet at $D$. Through the intersection of the two smaller circles at $J$ draw $DJ$ meeting the given circle at $E$. Let $HD$ intersect the circle about $H$ at $B$, and (changing here the order of construction) join $CB$.  Through $D$ draw $AM$ parallel to $GC$, and join $AC$. Prove that $B$ lies on $AC$ (cf. @Phil H), i.e. that $CB$ extended crosses the given circle at $A$, making $AD$ perpendicular to $DE$. 
But first it must be shown that $D$ is the center of the given circle.
Joining $GF$, $FH$, $HC$, and $KH$, then since $FG=HC$, $FGCH$ is an isosceles trapezoid with$$FH\parallel GC$$and by symmetry $DE$ perpendicularly bisects $GC$ and $FH$.

Again, since $GF=FH=HK$, then $GFHK$ is a rhombus, with $KG$, $KH$ making equal angles with diagonal $FK$, whence the center of the given circle lies on $FK$ extended. 
It follows by a like argument that the center of the circle lies on $HL$. Therefore $D$ is the center.
Now since $AM\parallel GC$, then$$\angle GCA=\angle MAC$$But (again cf. @Phil H)$$\angle KCB=\frac12\angle KHB=\frac12\angle ADG=\frac12\angle MDC=\angle MAC$$Therefore$$\angle GCA=\angle KCB$$$CB$ extended to the given circle at $A$ meets the line draw through center $D$ parallel to $GC$, and consequently$$AD\perp DE$$
