Orthogonal circles and chords

Question

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.

Answer 1

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.

Answer 2

Solution from a friend of mine (with a large contest-math geometry background).
$\angle OAC=\angle ADC$ by Tangent-secant theorem, hence $$\triangle OAD\sim \triangle OCA\Rightarrow \frac{AC}{AD}=\frac{OC}{OD}.$$ Similarly $$\frac{BC}{BD}=\frac{OC}{OD} \hbox{, hence }\frac{BC}{BD}=\frac{AC}{AD}\Rightarrow BC\cdot AD=AC\cdot BD\hbox{, QED.}$$