Can you provide a proof for the following claim:
Claim. Given two concentric circles $k_1$,$k_2$ and the third circle $k_3$ which intersects both $k_1$ and $k_2$ at the points $A$ and $B$ respectively. A point $P$ is selected on the circle $k_3$ on the outside arc. Let $C$ be projection of the point $P$ through point $A$ onto circle $k_1$ and $D$ projection of the point $P$ through point $B$ onto circle $k_2$. Then the length of the line segment $CD$ doesn't depend on the location of the point $P$ on its arc.
GeoGebra applet that demonstrates this claim can be found here.
In the case when radii of the circles $k_1$ and $k_2$ are equal we can prove the claim by observing that $\angle CPD$ doesn't depend on the position of the point $P$ and that its measure is defined by the difference of the arcs (on the circle $k_1$) $CD$ and $AB$. Since the arc $AB$ is fixed so is the arc $CD$ , hence the length of the line segment $CD$ is constant. But how to prove the claim in the case when the radii of the circles $k_1$ and $k_2$ are not equal?
Best Answer
Let the centre of the concentric circles be $O_{1}$ and the centre of circle $k_{3}$ be $O_{2}$. Draw $O_{1}C$ and $O_{1}D$. Let another random point on circle $k_{3}$ be $P'$. Let points $C'$ and $D'$ be the equivalent points for this point $P'$. Draw $PC'$, $PD'$, $C'D'$, $O_{1}C'$ and $O_{2}D'$.
Proving $CD$ and $C'D'$ equal is equivalent to proving $\angle CO_{1}D=\angle C'O_{1}D'$ and thereafter proving $\triangle CO_{1}D$ and $\triangle C'O_{1}D'$ congruent.
Observe that, $\angle CO_{1}C'=2\angle CAC'=2\angle PAP'=2\angle PBP'=2\left(180-\angle DBD'\right)=\angle D'O_{1}D$.
$\Rightarrow \angle CO_{1}C'=\angle D'O_{1}D$
Adding $\angle CO_{1}D'$ to both sides gives $\angle CO_{1}D=\angle C'O_{1}D'$.
In $\triangle C'O_{1}D'$ and $\triangle CO_{1}D$, $C'O_{1}=CO_{1}$, $D'O_{1}=DO_{1}$ and $\angle CO_{1}D=\angle C'O_{1}D'$. Hence, $\triangle C'O_{1}D'\cong \triangle CO_{1}D$ and therefore, $CD=C'D'=constant$.