Proving an approximate angle trisection by compass and straightedge My 70 year old father has given the below explanation. Can some one please verify or point out deficiencies if any.

The geometrical problem of trisecting any given angle by using compass and a straight unmarked edge is impossible. Many have tried but couldn't achieve the perfect one over a period of nearly two centuries. l hereby submit a near perfect but simple way of construction as under.

Given. In a circle with center $O$, $\angle POB$ is any angle to be trisected. $A$-$O$-$B$ is diameter extended towards $Q$.
Construction. Join $AP$. By construction, take $C$ as midpoint of $AP$. From $C$ we draw a transversal at an angle of one fourth of given $\angle POB$, which is possible by geometrical construction. (Eg, if the measure of $\angle POB$ is, say, $12x$, then $\angle PCR$ is of measure $3x$.) Our transversal meets a line parallel to diameter in $R$ and diameter extended in $Q$. Join $PQ$ and $RA$, which gives us a parallelogram $PRAQ$.
If you measure the $\angle PQO$ by protractor, you will find that it's measure is nearly one third of the measure of the $\angle POB$.

Proof. I have tried my level best to prove it geometrically but was not successful, however found the proof of impossibility.
Let this time the angle be $90^\circ$ (i.e., a right angle) and the radius of the circle be unity (i.e., one). Then $\angle PQO$ must be of $30^\circ$, which means segments $PQ$ and $QO$ must be $2$ and $\sqrt3$, respectively. Also, segments $MQ$ and $MO$ equal to $1$.
$\triangle APO$ is an isosceles right-angled triangle, measure of base angles are $45^\circ$, each which means segment $AP$ is $\sqrt2$. $C$ being midpoint of $AP$, segment $AC$ is $\sqrt2/2=1/\sqrt2$. Thus, all segments $PR$, $PC$, $AC$, and $AQ$ are $1/\sqrt2$ each. (We can prove this being opposite sides of parallelogram.) Therefore, the length of $QO$  is $AO+AQ=1+1/\sqrt2$. Certainly $\sqrt3$ is not equal to $1+1/\sqrt2$. thus the proof of impossibility.
 A: The construction can be simplified to this:

(To see why: $\angle BAP=\frac12\angle BOP$ by the Inscribed Angle Theorem. In the construction, $\overline{QC(R)}$ makes an angle of $\frac14\angle BOP=3\theta$ with $\overline{AP}$. Thus, $\triangle ACQ$ has an interior angle of $3\theta$ and an exterior angle of $6\theta$, so its other angle is $3\theta$: that is, it's isosceles with $\overline{AC}\cong\overline{AQ}$. Consequently, locating $Q$ is a simple matter of transferring $C$ to the extended diameter via a circle about $A$.)

To understand the nature of the approximation ...

By Thales' Theorem, $\angle APB$ is a right angle. Therefore, $|AP|=|AB|\cos6\theta$, so that $$|AQ|=|AC|=|CP|=r\cos 6\theta \tag1$$
If we drop a perpendicular from $P$ to $B$, we find from right triangle  $\triangle PDQ$ that
$$\tan\angle Q = \frac{r\sin 12\theta}{r\cos6\theta+r+r\cos12\theta} =
\frac{2\sin6\theta\cos6\theta}{\cos6\theta+2\cos^26\theta}= \frac{2\sin 6\theta}{1+2\cos6\theta} \tag{2}$$
Thus, the construction amounts to the assertion

$$\tan\frac13\phi \;\approx\;\frac{2\sin\frac12\phi}{1+2\cos\frac12\phi} \tag{3}$$

As series, this says
$$\color{green}{\frac13\phi}+\frac1{81}\phi^3+\frac2{3645}\phi^5+\cdots \;\approx\;
\color{green}{\frac13\phi}+\frac1{72}\phi^3+\frac{13}{17280}\phi^5+\cdots \tag{3'}$$
Also, as noted in a comment to the original question, we have
$$\cos\frac13\phi \;\approx\; \frac{1+2\cos\frac12\phi}{\sqrt{5+4\cos\frac12\phi}} \quad\to\quad
\color{green}{1 - \frac{\phi^2}{18}} + \frac{\phi^4}{1944} + \cdots\;\approx\;
\color{green}{1 - \frac{\phi^2}{18}} - \frac{\phi^6}{233280} -\cdots \tag{4}$$
(The reader can consider other forms.) So, all in all, the approximation doesn't seem terrible. A thorough error analysis is left as an exercise.
