# Proof for tangent chord angle formula

I want a proof for tangent chord angle formula by using the following method: Drawing a parallel line - See the diagram. I know the other proofs and I want to prove it with drawing a parallel line.

If $AB \parallel CD$, then $\angle BAC=\angle ACD$.
Because $AB$ is tangent (and $AB \parallel CD$), then $|AC|=|AD|$, so the triangle $ACD$ is isosceles and $\angle ADC=\angle ACD$.
$\angle AOC$ is central angle with inscribed angle $\angle ADC$, so $$(\widehat{AC}=)\angle AOC = 2 \angle ADC = 2 \angle ACD = 2 \angle BAC$$ We have then $$\frac{\widehat{AC}}{2}=\angle BAC$$
• Hi. How did you get the first part? I know one proof for the first part but it uses the fact that $\angle BAC = \widehat {AC} /2$ which is what we want to prove in the question. – titansarus Jan 13 '17 at 5:29