Right triangle $\triangle ABC$, $D$ lies on $AB$, inscribed in a circle of radius $9$. Find the measure of $CD$ I saw this mathematical puzzle on an Instagram post today. As title suggests, we have a right angled triangle inscribed in a circle with radius $D$ and some angles. The goal is to find the length of $CD$. I'll share my approach as an answer below. I'm not quite sure if my answer is correct, so please feel free to point out any faults in my approach and/or post your own approaches too!

 A: This is going to be my approach to the problem:

Here's my approach for the problem:
1.) Extend $CD$ to meet $E$ that lies on the circumference. Join the center of the circle $O$ with $E$ via segment $OE$. Since $OE$ is radius, we know that $OE=9$. Notice that $\angle ECB$ is the inscribed angle of $\angle EOC$, therefore we can conclude that $\angle EOB=2\alpha$
2.) Notice that $\angle EOA=\angle EDB=180-2\alpha$, therefore $OE=ED=9$. Now we can use the intersecting chords theorem (which can easily be proven via similarity of triangles in any general cyclic quadrilateral) and conclude that:
$$9\cdot CD=6\cdot12=72$$
$$\Rightarrow CD=8$$
A: Your solution is much nicer than what I write in the following. However, the solution below is almost straightforward.
Let's assume: $AC=y, BC=x$. Then:
$$\frac{x}{12}=\frac{\sin 2\alpha}{\sin \alpha} \implies \frac{x}{24}=\cos \alpha \\\frac{y}{6}=\frac{\sin (\pi - 2\alpha)}{\sin (\frac{\pi}{2}-\alpha)} \implies \frac{y}{12}=\sin \alpha.$$
Hence, $x^2+4y^2=24^2$. On the other hand, we already have :$x^2+y^2=18^2.$So,
$$y^2=84, x^2=240.$$
By the law of cosines in $\triangle ADC $, we obtain:
$$AD^2=36=y^2+CD^2-2CD\times y\cos (\frac{\pi}{2}-\alpha)=\\y^2+CD^2-2CD\times y\times \frac{y}{12}\implies \\ 48=14CD-CD^2\implies CD=8 \ or\ 6.$$
But $CD=6$ is not acceptable because that will imply that $\alpha= 45^{\circ}$, which is impossible.
A: Trig identity: $\displaystyle \;\frac{\sin 3α}{\sin α} - \frac{\cos 3α}{\cosα} 
= \frac{\sin 2α}{(\sin α)(\cos α)} = 2$
Law of Sine, ΔBCD:
$\displaystyle\frac{x}{\sin (\pi-3α)} = \frac{BC}{\sin 2α} = \frac{12}{\sin α}$
$→ \;\;\displaystyle \frac{x}{\left(\frac{\sin 3α}{\sin α}\right)} = \frac{BC}{2 \cos α} = 12$
Right triangle ΔABC:
$\;BC = 18\cos(\pi-3α) = -18\cos(3α)$
$\displaystyle \frac{BC}{2 \cos α} 
= -9 \left(\frac{\sin 3α}{\sin α} - 2\right)$
$\displaystyle → \quad 12 = -9 \left(\frac{x}{12} - 2\right) \qquad
→ \large\;CD = x = 8$
A: The fantastic power of a point wrt. a circle solution is already found by Goku but let me show my try:
By the law of cosines on $\triangle CDO$, $OC^2=CD^2+DO^2-2CDDO\cos2\alpha$. Putting $CD=x$, $DO=3$, $OC=9$ in this equation we have
$$4x^2-24x-288+48x\sin^2\alpha=0\tag1$$
On the other hand, by the law of sines on $\triangle CDB$, $\frac{x}{12}=\frac{\sin3\alpha}{\sin\alpha}$. By the identity $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$,
$$48x\sin^2\alpha=36x-x^2\tag 2$$
Combining $(1)$ and $(2)$, we have $x^2+4x-96=0$ and $x=8$ or $x=-12$. Since $x$ is positive, $x=8$.
A: A novel way, by calculating area 2 different ways.
Extend CD to intersect the circle at point E.  Let $\,x=CD\,,\, y=DE$
Area(ΔAEB), with base = EB
$\displaystyle \frac{(18\sin α)\,(18\cos α)}{2} = 81\,(\sin 2α)$
Area(ΔAEB), with base = AB
$\displaystyle \frac{18 × y\,\sin(\pi-2α)}{2} = 9y\,(\sin 2α)$
Area, done in 2 ways, must match: $\quad→y = 9$
$ΔCDB \sim ΔADE\;\;→ \displaystyle\frac{x}{6}=\frac{12}{y}\quad → x = 8$

Even better, Law of Sine on ΔADE:
$\displaystyle \frac{y}{\sin α} = \frac{18 \cos α}{\sin 2α}\;\; → y=9$
A: Law of Tangents, let $t = \tan α$
$\displaystyle \frac{x+6}{x-6} = 
\frac{\tan \frac{(3α-90°)+(90°-α)}{2}}{\tan \frac{(3α-90°)-(90°-α)}{2}}
= \frac{\tan α}{\tan (2α-90°)} = -(\tan α)(\tan 2α) = \frac{-2t^2}{1-t^2}$
$\displaystyle \frac{x+12}{x-12} = 
\frac{\tan \frac{(180°-3α)+(α)}{2}}{\tan \frac{(180°-3α)-(α)}{2}}
= \frac{\tan (90°-α)}{\tan (90°-2α)} = \frac{\tan 2α}{\tan α} = \frac{2}{1-t^2}$
$→ \displaystyle \frac{x+6}{x-6} + \frac{x+12}{x-12} = 2$
$→ x^2-6×12 = (x-6)(x-12) $
$\displaystyle → x = \frac{6×12}{(6+12)÷2} = \frac{72}{9} = 8$
