I have troubles to prove the following:
Let $\Gamma$ be a circle with center $O$, $a$ be a tangent to $\Gamma$, $A=a\cap \Gamma$, $D$ a point on $a$ and $B\in \Gamma$ such that $D$ and $B$ lies on the same side of the line $OA$. Prove that $\angle AOB=2\angle DAB$.
I was wondering whether the theorem in https://proofwiki.org/wiki/Angles_made_by_Chord_with_Tangent is useful but I think that in general case the line segment $AB$ won't go through the point $O$.
How can I prove the theorem?
$\triangle OAB$ is an isosceles triangle. If $\angle DAB = \theta$, then $\angle OAB = 90-\theta$, and you should be able to get $\angle AOB = 2\theta$ for there.