Proving segments equal $(O)$ is internally tangent to a circle $(O')$ at $F$ i.e $(O')$ is inside $(O)$. Let $CE$ be a chord of $(O)$ which touches $(O')$ at $D$. Let $CO$ meet $(O)$ again at $A$ and $CO'$ meet $AE$ at $B$. It is given that $OO' \perp AC$. Prove that $AB=CD$.

From working backwards and using power of point arguments and a bit of calculation,our goal is to prove that $AB^2=AC\times OO'$ which after some trigonometry implies $\sin^2\alpha=\sin 2\theta$ where $\alpha=\angle ABC$ and $\theta=\angle ACB$,but I can't proceed forward from here.
Also a synthetic or projective solution will be much appreciated.
 A: 
For brevity, we denote the radii of the smaller circle and the larger circle $r$ and $R$ respectively. By applying Pythagoras's theorem to the right-angled triangle $COO'$, it is possible to show,
$$CO'^2 = CO^2+OO'^2=R^2+\left(R-r\right)^2\quad\longrightarrow\quad CO'=\sqrt{2R^2+r^2-2Rr}. \tag{1}$$
To determine the length of $CD$, we apply the same theorem to the right-angled triangle $CDO'$.
$$CD^2=CO'^2 - O'D^2 = 2R^2+r^2-2Rr - r^2 \quad\longrightarrow\quad CD=\sqrt{2R\left(R-r\right)} \tag{2}$$
At this juncture, in order to facilitate our proof, we add the auxiliary line segment $O'D$ to the diagram given in the problem statement and extend it to meet $CA$ at $G$. Since $CD$ is the tangent to the given circle at $D$, $O'D$ is perpendicular to $CD$.
Since $\angle AEC$ is inscribed in a semicircle, $\measuredangle AEC = 90^o$. Therefore, $AB$ is parallel to $GO'$ and we shall write,
$$\dfrac{AB}{GO'}=\dfrac{CE}{CD}. \tag{3}$$
Since $\measuredangle CDO' = \measuredangle COO'=90^o$, $CODO'$ is a cyclic quadriateral. Therefore, $\measuredangle DCO = \measuredangle DO'O$. Hence, we can state that
the two right-angled triangles $CAE$ and $GO'O$ are similar. So, we have,
$$\dfrac{CA}{GO'}=\dfrac{CE}{OO'}. \tag{4}$$
We can use (2), (3), and (4) to show that
$$AB\times CD = CA\times OO'\quad\longrightarrow\quad AB=\dfrac{2R\left(R-r\right)}{\sqrt{2R\left(R-r\right)}}=\sqrt{2R\left(R-r\right)}=CD.$$
A: Here's a "horrible" proof using algebraic manipulations.

Set things in coordinate plane as in the picture. That is, $O$ is the unit circle centered at origin and center of $O'$ lies on the $y$-axis. Let the circle $O'$ have radius $r$. Denote the x-coordinates of points $D, E, B$ as $x_1, x_2, x_3$, respectively and let the line through $C$ and $D$ have slope $k$.
Let's find useful equations for the quantities. Using the fact that $D$ is a tangent point, we have $k = \frac{x_1}{\sqrt{r^2-x_1^2}}$ and $\frac{x_1}{\sqrt{r^2-x_1^2}}(x_1+1) = 1-r-\sqrt{r^2-x_1^2}$. These can be manipulated to
$$\begin{align*}(x_1+r^2)^2 &= (1-r)^2(r^2-x_1^2) \tag{1} \\
k &= \frac{x_1(1-r)}{x_1+r^2} \end{align*}
$$
Then solve $x_2$ from ($E$ on circle $O$ and the $y$-coord comes from the line equation)
$$x_2^2+k^2(x_2+1)^2=1$$
by factoring $k^2(x_2+1)^2=1-x_2^2 = (1-x_2)(1+x_2)$ and cancelling to get
$$x_2 = \frac{1-k^2}{1+k^2}.$$
Then for $x_3$, we need to solve an intersection of two lines. There's some nice cancelling and we get
$$x_3 = \frac{1-k(1-r)}{1+k(1-r)}.$$
Now we know the coordinates of the points $D$ and $B$ in terms of $r$ and $x_1$. And we have equation (1) to help us prove the equality of the lengths in question. Let's write out their squares
$$d(C,D)^2 = (1+x_1)^2 + k^2(x_1+1)^2 = (1+x_1)^2(1+k^2)$$
$$d(A,B)^2 = (1-x_3)^2 + ((1-r)(x_3+1))^2 = \left(\frac{2(1-r)}{1+k(1-r)}\right)^2(1+k^2).$$
So, it suffices to prove that
$$\frac{2(1-r)}{1+k(1-r)} = 1+x_1.$$
Plugging in $k=\frac{x_1(1-r)}{x_1+r^2}$ and using $(1-r)^2 = \frac{x_1+r^2}{r^2-x_1^2}$ we turn the LHS to $\frac{2(1-r)(r^2-x_1^2)}{r^2(1+x_1)}$.
Finally, multiplying by the denominator, it suffices to prove
$$2(1-r)(r^2-x_1^2) - r^2(1+x)^2 = 0,$$
which when multiplied out, can be seen to be equivalent to equation (1).
