Bisecting an angle doesn't lead to Trisecting? http://en.wikipedia.org/wiki/Talk:Angle_trisection
If you take the angle, and draw a circle at the corner of the angle. You mark two points along the edges of the angle.
Those two points form the tangent of an isosceles triangle.
If you can trisect a line, trisecting the tangent doesn't equal trisecting the angle?
 A: Let $ABC$ be an isosceles triangle, with $AB=AC$. Divide the segment $BC$ into $3$ equal parts, using trisection points $P$ and $Q$. So $AP=PQ=QB$.
Then $\angle BAP$ is not equal to $\angle PAQ$.
One way of showing this is to observe that the lines $AP$ and $AQ$ divide $\triangle ABC$ into $3$ triangles of equal area (equal bases, same height).  
But the area of $\triangle BAP$ is $\frac{1}{2}(AB)(AP)\sin(\angle BAP)$. 
Similarly, the area of $\triangle PAQ$ is $\frac{1}{2}(AP)(AQ)\sin(\angle PAQ)$.
Since $AB\gt AQ$, it follows that $\angle PAQ$ is bigger than angle $BAP$. 
A: Here is a concrete example of why this fails.  Take an isosceles right triangle, with both arms of length 1, and diagonal of length $\sqrt{2}$.  Divide the diagonal into thirds, and you get the following picture:
Now, side $A$ has length $\frac{\sqrt{2}}{3}$, since it's one-third of the total diagonal, and side $B$ has length $1$.  Angle $c$ is $45^\circ$, and $a+b+c=180^\circ$.  We now apply the law of sines to get $$ \frac{\sin a}{A}=\frac{\sin b}{B}$$
which simplifies to $$\sqrt{2}\sin b=3\sin a ~~~~~~~(\star)$$  But now we use the angle-sum formula for sine to write $$\sin b=\sin(180^\circ-45^\circ-a)=\sin(135^\circ-a)=\sin 135^\circ\cos a - \cos 135^\circ \sin a$$
Since $\sin 135^\circ=\frac{1}{\sqrt{2}}$ and $\cos 135^\circ=-\frac{1}{\sqrt{2}}$, we plug  this into $(\star)$ to get $$\cos a +\sin a = 3\sin a$$
Now, we subtract $\sin a$ from both sides, and divide to get $$\frac{\sin a}{\cos a}=\frac{1}{2}$$
Hence $a=\tan^{-1} \frac{1}{2}\approx 26.6^\circ$, which isn't $30^\circ$, a third of $90^\circ$.
A: Suppose that we want to "trisect" an angle $3\theta$ by the method suggested.  By using basic trigonometry and then calculating a Taylor series we find that the middle of the three angles constructed is
$$2\tan^{-1}\Bigl(\frac{1}{3}\tan\frac{3\theta}{2}\Bigr)
  =\theta+\frac{2}{3}\theta^3+\frac{1}{2}\theta^5+\cdots\ .$$
This is not equal to $\theta$, so the trisection is not exact.
However, this gives some kind of answer to the comment by David H on Andre Nicolas' answer: the error in the trisection is approximately proportional to $\theta^3$, so for small values of $\theta$, the construction will be very close.
