euclidean-geometrygeometryratio
Two circles cut orthogonally in $A$ and $B$ . A diameter of one of the circles is drawn cutting the other circle in $C$ and $D$. Prove that $BC \cdot AD = AC\cdot BD$.
I have attempted a proof based on similar triangles and the usual angle theorems such as alternate segment,but can’t complete the proof.Any help would be appreciated.
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Let $D,E,F$ be a point on $BC,CA,AB$ such that $AD\perp BC,BE\perp AC,CF\perp AB$ respectively.
Then, we know that $E,F$ exist on the two circles.
Now consider the tangent lines at $F$ for both circles.
Then, let $G$ be a point both on the tangent line for the circle on $AH$ and on $BC$. Also, let $I$ be a point both on the tangent line for the circle on $BC$ and on $AC$.
Now we have $$\angle{CFG}=\angle{FAH},\quad \angle{IFC}=\angle{FBC}$$
Since $$\angle{FAH}+\angle{FBC}=\angle{BAD}+\angle{ABD}=90^\circ$$ we have $$\angle{CFG}+\angle{IFC}=90^\circ.$$
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.
Best Answer
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.