euclidean-geometrygeometry
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.
Mine look like this. If $\angle CAB=\theta$ then $\angle DAB=180-\theta$ and, $\angle S_{1}CB=90-\theta\rightarrow\angle MCB=180-\theta$ , $\angle S_{2}DB=90+\theta\rightarrow \angle MDB=\theta$.
$\angle MCB + \angle MDB = 180$ implies $MCBD$ is a cyclic quadrilateral
Orthogonality means that $OB\bot BO'$ so $OB$ is tangent to circle with center at $O'$.
Because of tangent chord property we see that (blue) $$\Delta BCO\sim \Delta DBO\implies {OB\over OD} = {BC\over BD}$$ Similary (red) we have $$\Delta ACO\sim \Delta DAO\implies {OA\over OD} = {AC\over AD}$$
Since $OA = OB$ we are done.
Best Answer
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.