Given a segment $r$, a segment $c$, and an angle $\angle AOB$, construct (straightedge and compass) a circle whose center is on ray $OA$, whose radius is congruent to $r$, and where the chord created by ray $OB$ is congruent to $c$.
Source: Based on Kiselev's Geometry
My solution is below. It seems too complicated than I would expect, so I ask:
- Is it correct?
- Is there a simpler solution?
Background: See here for background. Thank you to X-Rui for helping me out.
Solution: Let segment $c$ have endpoints $U$ and $V$. Draw two circles with radius $r$ and centers $U, V$ respectively and label one of their intersections $X$. Drop the perpendicular from $X$ to $UV$ and label its foot $T$.
From $O$, erect segment $OC$ perpendicular to $OB$ with length congruent to $TX$. From $C$, draw ray $CD$ parallel to $OB$ and label its intersection with $OA$ as $D$.
Now draw circle $\bigcirc D$ with radius $r$. Label the intersections of $\bigcirc D$ with $OA$ as $F$ and $G$. Then chord $FG$ is congruent to $UV$, completing the construction.
Proof: Drop the perpendicular from $D$ to $OB$ and let its foot be $H$. Segment $HD \cong OC \cong TX$. Likewise, $DF \cong DG \cong XU \cong XV$. Therefore $FG \cong UV$ and $\bigcirc D$ is the desired circle.