Is it possible, given a point in the interior of an angle to construct a circle through that point tangent to the rays forming the angle?
Suppose you have a point $A$ in the interior of an angle with vertex $O$. You can bisect the angle, take a point $I$ on the angle bisector, and drop perpendiculars $IP$ and $IQ$ to each side of the angle. The circle with center $I$ and radius $IP$ is then tangent to the rays of the angle. Call this circle $\Gamma$.
You can then draw a line $OA$, which will intersect $\Gamma$ at some point $A'$ on the far side of $\Gamma$ from $O$. You can then draw lines $EA$ parallel to $PA'$ and $FA$ parallel to $QA'$, where $E$ is the intersection with the ray containing $P$, and $F$ is the intersection with the ray containing $Q$. It seems intuitively obvious that the circle through $EAF$ will be a circle through $A$ tangent to the rays of the angle. But is there a way to prove it? I don't see how to do so. Thanks.