We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
78
2
avatar+42 

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 $\angle YZX$?

 May 16, 2019
 #1
avatar+8406 
+4

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 \(\angle YZX\)?

----------

 

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°

 May 16, 2019
 #2
avatar+42 
+2

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

Thanks

 May 20, 2019

8 Online Users

avatar