Constructing the circle inversion inverse of a point with ruler only I've been reading a bit about inversive geometry, particularly circle inversion. The following is a problem from Hartshorne's classical geometry, which I've been struggling with on and off for a few days.

I figured it would be helpful to show that $TU\perp OA$ first. At best, I tried to label angles according to which ones are congruent with each other. I know $\angle RPS$ and $\angle RQS$ are both right, as they subtend the diameter, so $\angle TPS=\angle UQS$ are both right as well. So $PTUQ$ is a cyclic quadrilateral, and thus $\angle RTQ=\angle PUQ$. Labeling $\angle SPQ$ as $3$ and $\angle PQS$ as $4$, I see that $1+2+3+4$ sum to two right angles. 
That's about as far as my observations got me. My hunch is that $RTU$ is an isosceles triangle, and $PU$ is like a line of symmetry, but I'm not sure how to show it, and how to eventually conclude $TU$ meets $OA$ at $A'$, that is, $OA\cdot OA'=r^2$, where $r$ is the radius if $\Gamma$. Thanks for any ideas on how to solve this.
 A: The three perpendiculars from the corners of a triangle to the sides meet in a common point, the orthocentre. You know that $UP$ is perpendicular to $RT$ and $TQ$ is perpendicular to $RU$; it follows that the line through $R$ and their intersection $S$ is perpendicular to $TU$.
[Update:] It turns out you can actually go on deducing all the angles much like you started. I'll use your angles $1$, $2$ and $3$. (I don't understand how you defined $4$; in the drawing it seems to mark the right angles $\angle SPT$ and $\angle SQU$ but in the text you defined it as $\angle PQS$.)
First, to make the symmetry of the situation manifest, let's also draw the lines $PA'$ and $QA'$. Here's an image (I'll be justifying the angles I filled in in a bit):

The triangle $PQA'$ is the orthic triangle of the triangle $RTU$. If you take out the circle $\Gamma$, the diagram has $S_3$ symmetry, so the quadrilaterals $A'URP$ and $QRTA'$ are cyclic for the same reason as $PTUQ$; that justfies the angles I've filled in. Since $|OR|=|OP|=r$, triangle $ROP$ is isosceles, so $\angle OPR$ is 3, and hence $\angle OPA$ is 1. Thus the triangles $OPA$ and $OPA'$ have two angles in common (1 at $\angle OPA$ and $\angle OA'P$, and $\angle POA=\angle POA'$), and hence are similar. The inversion property then follows by taking the ratios of corresponding sides in these triangles.
P.S.: That the altitudes of $RTU$ are the bisectors of its orthic triangle is related to the fact (mentioned in the Wikipedia article) that the orthocentre is the incentre of the orthic triangle.
A: i will follow up on jorikis answer. first we will show that A and A' cut the diagonal RS harmonically. that is the cross ratio of four collinear points $(R, S; A, A^\prime)$ defined by $\frac{RA}{SA} / \frac{RA^\prime}{SA^\prime}$ is unity. this will imply that $A$ and $A^\prime$ are conjugates. cross ratio $(R, S; A, A^\prime)$ is equal to the cross ratio
of the lines $(RP, SP; AP, A^\prime P)$ which is 
$\frac{\sin(\angle RPA) \sin(\angle S P A^\prime)}{\sin(\angle SPA) \sin(\angle RPA^\prime)}$ by law of sine. this is unity because $\angle RPS = 90^\circ$ and the line $PS$ bisects 
$\angle AP A^\prime.$
p.s. i got this idea by picking $P$ so that $AP$ is orthogonal to $RS.$ in this special case
the result follows easily. now generalize. 
