Points $A$ and $C$ with position vectors $u$ and $v$ lie on the circle with centre $0$ and radius $r$. Tangents to the circle at $A$ and $C$ meet at the point $B$ with position vector $w$.
Show that $||w-u||=||w-v||.$
I have attempted this algebracilly reaching: $\sqrt{<w,w>-2r^2+<u,u>}=\sqrt{<w,w>-2r^2+<v,v>}$
From here can I just simply say $<u,u>$ and $<v,v>$ are both $r^2$ because they are both lines from the origin to the perimiter? This is what I have done and have ended up with:
$<w,w>=<w,w>$ where I conclude that $||w-u||=||w-v||.$
However I feel like this may not have been right to assume because if I do that surely I could just sub in $u=v=r$ right at the start which would defeat the object of any of this. Is there a better method to do this?
Observe $ \Delta ABO and \Delta CBO $
$BO = BO$ (Common side)
Angle at $A$ and $C$ is equal to $90^0$
$AO = CO$ (Radius of circle)
Hence they are congruent. Hence $AB = BC$.
Now $ ||w-u|| = AB$ and $||w-v|| = BC$ and they are equal as proven above.