Circle $\Gamma$

Circle $\Gamma$ is the incircle of $\triangle ABC$ and is also the circumcircle of $\triangle XYZ$. The point $X$ is on $\overline{BC}$, the point $Y$ is on $\overline{AB}$, and the point $Z$ is on $\overline{AC}$. If $\angle A=40^\circ$, $\angle B=60^\circ$, and $\angle C=80^\circ$, what is the measure of $\angle YZX$?

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Instead of calling the center of the circle  $\Gamma$ , let's call it  G .

Y is a point on both the circle and its tangent line AB. So Y must be a point of tangency. The same goes for X.

Draw a line from G to Y and line from G to X. These form right angles with the tangent lines.

Look at quadrilateral BYGX. The sum of the angle measures in a quadrilateral  =  360° . So...

m∠YGX  =  360° - 90° - 90° - 60°  =  120°

The measure of an inscribed angle is half the measure of the central angle with the same intercepted arc.

m∠YZX  =  (1/2)(m∠YGX)  =  (1/2)(120°)  =  60°

Ohhhhhhh, I never thought of it like that. That is a brilliant way of solving that!

Thanks