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We have a $\triangle ABC$ and a circumscribed circle $k$. Line $c$ is parallel to the tangent of the circle in $C$. Show that $\angle CAB = \angle CA_1B_1$.

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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?

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    $\begingroup$ $\angle BAC=\angle BCM$ (inscribed angles on the same arc). $\endgroup$ – Intelligenti pauca Aug 15 '19 at 21:31
  • $\begingroup$ There is also another argument for the @Aretino 's statement: Let $T$ be the other intersection point of the line $OC$ and the circumscribed circle. Notice that $\measuredangle TBC=90^\circ$ and, since, $\measuredangle MCB$ and $CTB$ are of the same kind (both acute/right/obtuse), then $CT\perp CM\ \land\ BT\perp BC\implies\measuredangle MCB=\measuredangle CTB=\measuredangle CAB$ $\endgroup$ – Invisible Jun 28 '20 at 13:39
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$\angle CAB = \angle BCM$, 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, $\angle CAB = \angle CA_1B_1$.

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  • $\begingroup$ Oh, silly me! Thank you! $\endgroup$ – Kite Aug 15 '19 at 21:53
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Because $$\measuredangle CAB=\measuredangle MCB=\measuredangle CA_1B_1.$$

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