A problem in geometry(tangents) to show that $CD= 2AB \cdot (AC^2−AB^2)/(AC^2+AB^2)$

Question

Suppose that $O$ is the circumcentre of Triangle $BCD$. Draw the tangent to the circumcircle of $BCD$ at $B$. Extend $CO$ meet this tangent at $A$. Show that

$$CD= 2AB \,\frac{AC^2−AB^2}{AC^2+AB^2}$$

What I thought was to apply Cevas theorem or Menelaus theorem. But things are not working out.I need some little help. I can't show what I have done as absolutely other than construction I could not do anything. So I would be obliged if anyone can help.

Answer 1

Comment: We have:

$AC^2=AB^2+AC^2-2AB\cdot AC cos(\angle BAC)$

$CD^2=AC^2+AD^2-2 AC\cdot AD cos(\angle CAD)$

Now applying Simson and Steward theorems may help you to find the relation.

Answer 2

Showing why this is not necessarily true.

$\angle OCB = \angle OBC = \theta$

In $\triangle ABC,$

$\displaystyle \frac{AB}{\sin \theta} = \frac{AC}{\sin (90^0 + \theta)} = t$

$AB = t \sin \theta, AC = t \cos \theta$

That gives us

$ \frac {AC^2 − AB^2}{AC^2+AB^2} = \cos 2 \theta$

Now please note $\angle D = \frac{1}{2} \angle BOC = 90^0 - \theta$

Also $AB = R \tan 2 \theta \,$ (as $\angle AOB = 2 \theta \, , \,R$ is the circumradius)

$2 AB \cos 2 \theta = 2R \sin 2 \theta = 4 R \sin \theta \cos \theta = 4 R \sin D \cos D = 2 BC \cos D $