Chord in Orthogonal Circles Two orthogonal circles with centres $A$ and $B$ have a common chord which meets $AB$ at $C$. $DE$ is a chord of the first circle that passes through $B$. Prove that $A,D,E,$ and $C$ are concyclic.
I have attempted this question using angles in the alternate segment, and properties of cyclic quadrilaterals including Ptolemy's Theorem, but cannot proceed to a proof. Any help would be appreciated.
 A: Let $C'$ be the point of intersection of the circle passing through $A$, $D$ and $E$ with the line $AB$. We need to show that $C=C'$.
Let's first notice that the point $C'$ lies between points $A$ and $B$.
Let $D$ be the point closer to $B$ than $E$, so that  points $B$, $D$, $E$ lie on the line in that order.
Let
$$R = |AD| =|AE|$$
be the radius of the circle with the center $A$.
Let
$$\alpha = |\sphericalangle C'AD| = |\sphericalangle C'ED|$$
$$\beta = |\sphericalangle AC'E| = |\sphericalangle ADE|$$
(All of these angles are inscribed angles, so their measures are equal if they are subtended by the same arcs.)
Because $\triangle ADE$ is an isosceles triangle, it means that also
$$ |\sphericalangle AED| = |\sphericalangle ADE| = \beta$$
Since $\sphericalangle AC'D$ and $\sphericalangle AED$ are opposite angles in a quadrilateral inscribed in a circle, we have
$$ |\sphericalangle AC'D| = \pi - |\sphericalangle AED| = \pi-\beta $$
We also have
$$ |\sphericalangle ADB| = \pi- |\sphericalangle ADE| = \pi - \beta$$
To sum up, we have
$$ |\sphericalangle BAD| = |\sphericalangle C'AD| = \alpha$$
$$ |\sphericalangle ADB| = \pi-\beta = |\sphericalangle AC'D|$$
which means that $\triangle ABD$ and $\triangle ADC'$ are similar. From this we have
$$ \frac{|AB|}{|AD|} = \frac{|AD|}{|AC'|} $$
$$ |AC'| =\frac{|AD|^2}{|AB|} = \frac{R^2}{|AB|} $$
Independently it's not difficutlt to show that
$|AC| = \frac{R^2}{|AB|}$
(but tell me if you need help with this).
Since $|AC|=|AC'|$, and both points $C$ and $C'$ lie on the line between points $A$ and $B$, it means that $C=C'$, which finalizes the proof.
A: 
Proving $ACDE$ cyclic is equivalent to proving $\triangle BCD\sim \triangle BEA$ which is then equivalent to proving $BC\cdot BA=BD\cdot BE$.
To get $BC\cdot BA$, use the similarity of $\triangle BCQ$ and $\triangle BQA$.
This will give $BC\cdot BA=BQ^{2}$
Now, considering the power of point $B$ with respect to the circle with centre $A$ will give the value of $BD\cdot BE$ to be $BQ^{2}$.
Hence, $BC\cdot BA=BQ^{2}=BD\cdot BE$
A: Hint
Observe, by circle theorems or other methods that:
$$\bigtriangleup ADB \; \sim \; \bigtriangleup ECB$$
If the opposite angles add up to $180°$ in a quadrilateral, it is sufficient to conclude that it is cyclic.
For example, in the figure, let $\measuredangle CEB=\alpha$, then, $\measuredangle CED=180°-\alpha$ and $\measuredangle DAB + \measuredangle CED=180°$, confirming the $ADEC$ is cyclic.

A: 
Here’s an alternate proof that avoids trigonometry and establishes a neat bonus result that the circle ACDE is orthogonal to the second circle.
