Fix a circle $\bigcirc O$ and a diameter $d$. For each point $A$ on $\bigcirc O$, let $A'$ be its projection onto $d$, and mark $\Lambda$ on ray $OA$ such that $O\Lambda \cong AA'$. What is the locus of all such points $\Lambda$?
Source: Hadamard's Geometry
My proof is below, to which I request verification, critique, or alternatives.
Solution: Let diameter $d'$ be perpendicular to $d$, and mark the intersections of $d'$ with $\bigcirc O$ as $Y$ and $Y'$. Then the desired locus is the two circles $\omega, \omega'$ with diameters $OY$ and $OY'$.
Proof (Synthetic): Suppose $A$ is on the same side of $d$ as $Y$. Then we have $$O\Lambda \cong AA' \text{ (by construction)} \\ OY \cong OA \text{ (as both are radii of} \bigcirc O \text{)} \\ \angle \Lambda O Y \cong \angle A'AO \text{ (since } OY \parallel A'A\text{)}$$ so by SAS $$\triangle OY\Lambda \cong \triangle AOA'$$ and $\angle O \Lambda Y$ is a right angle.
Then by Thales' Theorem, $\Lambda$ is on circle $\omega$.
A similar argument shows that when $A$ is on the other side of $d$ then $\Lambda$ is on $\omega'$.
Conversely, if $\Lambda$ is any point on $\omega$, then by AAS, $\triangle OY\Lambda \cong \triangle AOA'$, so $O \Lambda \cong AA'$, completing the proof.
Proof (Analytic): Using $O$ as our origin, the desired locus clearly has the equation $$r = |\sin \theta|$$ for $\theta \in [0, 2 \pi]$. The circle $\omega$ has the equation $$\sqrt{(r\cos \theta)^2 + (r \sin \theta - \frac 1 2)^2} = \frac 1 2$$ for $r \geq 0, \theta \in [0, \pi]$, which simplifies to $$%\sqrt{r^2\cos^2 \theta + r^2 \sin^2 \theta -r\sin \theta + \frac 1 4} = \frac 1 2\\ \sqrt{r^2 -r\sin \theta + \frac 1 4} = \frac 1 2$$ and is clearly met if and only if $r = | \sin \theta|$ and $\theta \in [0, \pi]$. A similar argument applies to $\omega'$ when $\theta \in [\pi, 2\pi]$, completing the proof.