Geometry, Circles and Chords

Question

Two circles intersect at points $A$ and $B$. Chords $AC$ and $AD$ are drawn through the point $A$. Prove that $AC^2\cdot BD=AD^2\cdot BC$.

So, most probably, we'll use Power of point on $A$ and $B$, giving $AC^2=AB \cdot BC$ and $AD^2=AB \cdot BD$. So, finally, we get the desired result. Is this solution proper?

Thanks!!!

Answer 1

If chords $AC$ and $AD$ are tangents to another circles, so $$\Delta ABC\sim \Delta DBA, $$ which gives $$\frac{AB}{DB}=\frac{AC}{DA}=\frac{BC}{BA}$$ and from here we obtain: $$AB^2=BC\cdot BD,$$ $$\frac{\sqrt{BC\cdot BD}}{BD}=\frac{AC}{AD},$$ which is $$AC^2\cdot BD=AD^2\cdot BC,$$ which is exactly that you want to get.

Answer 2

I must be missing something, because it seems to me like C can still be made arbitrarily close to A, making AC arbitrarily close to zero, while neither AD nor BC approach zero. Does my picture represent a possible scenario for this problem, or am I missing something? (https://i.stack.imgur.com/ObtrA.jpg)