We have a $\triangle ABC$ and a circumscribed circle $k$. Line $c$ is
parallel to the tangent of the circle in $C$. Show that $\angle CAB = \angle CA_1B_1$.
So, $\angle CAB = \dfrac{\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{BQ} + \newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{QC}}{2}$ and $\angle CA_1B_1=\dfrac{\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{BQ} +\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{PC}}{2}$. From here, we see that we should prove that $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{QC}$ is equal to $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{PC}$.
The only way that I see is with congruent triangles ($OC\cap PQ=K; \triangle KPC \cong \triangle KQC$). Can we do it faster?
Best Answer
$\angle CAB = \angle BCM$, because they both inscribe $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{BC}$. And $\angle CA_1B_1 = \angle BCM$, because they are alternate interior angles of two parallel lines crossed by $BC$. Therefore, $\angle CAB = \angle CA_1B_1$.