IRL ABooklet

As $ABC$ is isosceles and $\angle ACB = 90^\circ$ then it follows that
$$\angle CAB = \angle CBA = 45^\circ$$
Now quadrilateral $ABEC$ is cyclic, so
$$\angle BEC = 180^\circ - \angle BAC = 135^\circ$$
Therefore by subtraction from $180^\circ$, we know $\angle BED = 45^\circ$.
Let us define $\theta = \angle BDE$. As we now have two angles of $\triangle BDE$, the third must take the total to $180^\circ$ and so
$$\angle EBD = 180^\circ - 45^\circ - \theta = 135^\circ - \theta$$
By subtracting $\angle EBD$ and $\angle CBA = 45^\circ$ from $180^\circ$, we deduce that:
$$\angle EBC = \theta.$$
Now as $\angle EBC$ and $\angle EFC$ are subtended by the common chord $EC$, we deduce that those two angles are equal. Putting these together, we deduce that $\angle BDE = \angle CFE$.
We can conclude that $\triangle CFE \sim \triangle CDG$ since $\angle ECF$ is common. Therefore,
$$\frac{|CE|}{|EF|} = \frac{|CG|}{|DG|}$$
which is what had to be proved.

Solution 2:

Connect $O$ to $C$ and $E$ to create the central angle $\angle EOC$, which is twice the angle $\angle EFC$. Because triangle $EOC$ is isosceles, we have
$$180^\circ = \angle EOC + 2\angle OCD = 2(\angle EFC + \angle OCD).$$
This implies $\angle EFC + \angle OCD = 90^\circ$, i.e. $\angle EFC = 90^\circ - \angle OCD$.
Because $AB$ is a diameter of the circle and triangle $ABC$ is isosceles, $CO$ is perpendicular to $AB$ and so $\angle CDO = 90^\circ - \angle OCD = \angle EFC$.
Because triangles $CFE$ and $CDG$ share the angle at $C$ and $\angle CDO = \angle EFC$ it follows that the two triangles $\triangle CFE$ and $\triangle CDG$ are similar. Therefore,
$$\frac{|CE|}{|EF|} = \frac{|CG|}{|GD|},$$
which gives the required equation $|CE| \cdot |DG| = |EF| \cdot |CG|$.

Solution 3:

(using trigonometry)
We use coordinates such that $O = (0,0)$, $C = (0,1)$, $A = (-1,0)$ and $B = (1,0)$. Let $\alpha = \angle BOE$ and $\beta = \angle BOF$, both in $[0, 2\pi)$. Because the points $E$ and $F$ are on the unit circle, these points have coordinates $E = (\cos \alpha, \sin \alpha)$ and $F = (\cos \beta, \sin \beta)$.
To determine the coordinates of $D$, we use the Intercept Theorem (or similar triangles) applied to the figure obtained by dropping the perpendicular $EP$ from $E$ onto the line $CO$ (the $y$-axis). This gives $|DO|/|OC| = |EP|/|PC|$ and so $|DO| = \frac{\cos \alpha}{1-\sin \alpha}$. Similarly, we obtain $|GO| = \frac{\cos \beta}{1-\sin \beta}$. If we work with signed distances, these formulas are correct for all positions of $E$ and $F$, provided that neither $E$ nor $F$ is equal to $C = (0,1)$.

Therefore,
$$\begin{align*} |DG|^2 &= \left( \frac{\cos \alpha}{1 - \sin \alpha} - \frac{\cos \beta}{1 - \sin \beta} \right)^2 \\ &= \frac{\cos^2 \alpha}{(1 - \sin \alpha)^2} + \frac{\cos^2 \beta}{(1 - \sin \beta)^2} - \frac{2 \cos \alpha \cos \beta}{(1 - \sin \alpha)(1 - \sin \beta)} \\ &= \frac{1 - \sin^2 \alpha}{(1 - \sin \alpha)^2} + \frac{1 - \sin^2 \beta}{(1 - \sin \beta)^2} - \frac{2 \cos \alpha \cos \beta}{(1 - \sin \alpha)(1 - \sin \beta)} \\ &= \frac{1 + \sin \alpha}{1 - \sin \alpha} + \frac{1 + \sin \beta}{1 - \sin \beta} - \frac{2 \cos \alpha \cos \beta}{(1 - \sin \alpha)(1 - \sin \beta)} \\ &= \frac{(1 + \sin \alpha)(1 - \sin \beta) + (1 - \sin \alpha)(1 + \sin \beta) - 2 \cos \alpha \cos \beta}{(1 - \sin \alpha)(1 - \sin \beta)} \\ &= \frac{2 - 2 \sin \alpha \sin \beta - 2 \cos \alpha \cos \beta}{(1 - \sin \alpha)(1 - \sin \beta)} \\ &= \frac{2(1 - \cos(\beta - \alpha))}{(1 - \sin \alpha)(1 - \sin \beta)}. \end{align*}$$

Since $\angle EOF = \beta - \alpha$, the Cosine Rule applied to triangle $EOF$ gives
$$|EF|^2 = 2(1 - \cos(\beta - \alpha)).$$
Similarly, we obtain
$$|CE|^2 = 2 \left(1 - \cos\left(\frac{\pi}{2} - \alpha\right)\right) = 2(1 - \sin\alpha).$$
Finally, the Theorem of Pythagoras on $\triangle COG$ gives
$$\begin{aligned} |CG|^2 &= 1 + |OG|^2 = 1 + \left( \frac{\cos \beta}{1 - \sin \beta} \right)^2 = \frac{(1 - \sin \beta)^2 + \cos^2 \beta}{(1 - \sin \beta)^2} \\ &= \frac{2}{1 - \sin \beta}. \end{aligned}$$
We now obtain
$$\begin{aligned} |CE|^2 \cdot |DG|^2 &= \frac{2(1 - \sin \alpha) \cdot 2(1 - \cos(\beta - \alpha))}{(1 - \sin \alpha)(1 - \sin \beta)} \\ &= 2(1 - \cos(\beta - \alpha)) \cdot \frac{2}{1 - \sin \beta} = |EF|^2 \cdot |CG|^2 \end{aligned}$$

We use complex numbers such that the circle is given by the equation $|z| = 1$, $C$ is the complex number $i$, $A = -1$ and $B = 1$. Let $E$ be $e^{i\alpha}$ and $F$ be $e^{i\beta}$. We determine the distances as functions of $t$ and $u$ where
$$t = \tan\left(\frac{\alpha}{2}\right) \quad \text{and} \quad u = \tan\left(\frac{\beta}{2}\right).$$
To aid calculations below, we observe
$$1 + t^2 = \frac{1}{\cos^2\left(\frac{\alpha}{2}\right)} \quad \text{and} \quad 1 + u^2 = \frac{1}{\cos^2\left(\frac{\beta}{2}\right)},$$
$$1 - \sin \alpha = 1 - 2 \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right) = \left(\cos \left(\frac{\alpha}{2}\right) - \sin \left(\frac{\alpha}{2}\right)\right)^2 = \frac{(1 - t)^2}{1 + t^2},$$
$$\cos \alpha = \cos^2 \left(\frac{\alpha}{2}\right) - \sin^2 \left(\frac{\alpha}{2}\right) = \frac{1 - t^2}{1 + t^2},$$
$$\begin{aligned} 1 - \cos(\alpha - \beta) &= 1 - \cos^2\left(\frac{\alpha - \beta}{2}\right) + \sin^2\left(\frac{\alpha - \beta}{2}\right) = 2\sin^2\left(\frac{\alpha - \beta}{2}\right) \\ &= 2\left(\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) - \cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\right)^2 \\ &= \frac{2(t - u)^2}{(1 + t^2)(1 + u^2)}. \end{aligned}$$

Because $D$ is on the line $CE$, it is of the form $\lambda e^{i\alpha} + (1-\lambda)i$ and has imaginary part equal to 0. Hence $\lambda \sin \alpha + (1 - \lambda) = 0$ and so $\lambda = \frac{1}{1 - \sin \alpha}$. This gives
$$D = \frac{\cos \alpha}{1 - \sin \alpha} = \frac{1 - t^2}{(1 - t)^2} = \frac{1 + t}{1 - t}.$$
A similar argument shows that $G = \frac{1+u}{1-u}$. We can now calculate the quantities $|CE|^2$, $|EF|^2$, $|CG|^2$, $|DG|^2$ as follows:
$$|CE|^2 = \cos^2 \alpha + (1 - \sin \alpha)^2 = 2(1 - \sin \alpha) = \frac{2(1 - t)^2}{1 + t^2}$$
$$|DG|^2 = \left| \frac{1+t}{1-t} - \frac{1+u}{1-u} \right|^2 = \frac{4(t-u)^2}{(1-t)^2(1-u)^2}$$
$$\begin{align*} |EF|^2 &= (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 = 2 - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta \\ &= 2(1 - \cos(\alpha - \beta)) = \frac{4(t-u)^2}{(1+t^2)(1+u^2)} \end{align*}$$
$$|CG|^2 = 1 + \frac{(1+u)^2}{(1-u)^2} = \frac{2(1+u^2)}{(1-u)^2}$$
Hence
$$|CE|^2 \cdot |DG|^2 = \frac{8(1-t)^2(t-u)^2}{(1+t^2)(1-t)^2(1-u)^2} \quad \text{and}$$
$$|EF|^2 \cdot |CG|^2 = \frac{8(t-u)^2(1+u^2)}{(1+t^2)(1+u^2)(1-u)^2}$$
and it follows that $|CE| \cdot |DG| = |EF| \cdot |CG|$.