# Show that angles are equal in a circumscribed circle

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, $$\angle CAB = \dfrac{\newcommand{arc}{\stackrel{\Large\frown}{#1}}\arc{BQ} + \newcommand{arc}{\stackrel{\Large\frown}{#1}}\arc{QC}}{2}$$ and $$\angle CA_1B_1=\dfrac{\newcommand{arc}{\stackrel{\Large\frown}{#1}}\arc{BQ} +\newcommand{arc}{\stackrel{\Large\frown}{#1}}\arc{PC}}{2}$$. From here, we see that we should prove that $$\newcommand{arc}{\stackrel{\Large\frown}{#1}}\arc{QC}$$ is equal to $$\newcommand{arc}{\stackrel{\Large\frown}{#1}}\arc{PC}$$.

The only way that I see is with congruent triangles ($$OC\cap PQ=K; \triangle KPC \cong \triangle KQC$$). Can we do it faster?

• $\angle BAC=\angle BCM$ (inscribed angles on the same arc). – Intelligenti pauca Aug 15 '19 at 21:31
• 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 $CT\perp CM\ \land\ BT\perp BC\implies\measuredangle MCB=\measuredangle CTB=\measuredangle CAB$ – Invisible Jun 28 '20 at 13:39

$$\angle CAB = \angle BCM$$, because they both inscribe $$\newcommand{arc}{\stackrel{\Large\frown}{#1}}\arc{BC}$$. And $$\angle CA_1B_1 = \angle BCM$$, because they are alternate interior angles of two parallel lines crossed by $$BC$$. Therefore, $$\angle CAB = \angle CA_1B_1$$.
Because $$\measuredangle CAB=\measuredangle MCB=\measuredangle CA_1B_1.$$