Question 5

In triangle $$ABC$$, we are given that $$\angle CAB = 80^\circ$$. Let the perpendicular bisector of $$BC$$ meet the circumcircle of triangle $$ABC$$ in $$N$$, where we assume that $$A$$ and $$N$$ lie on the same side of the chord $$BC$$. Then what is the measure of $$\angle NBC$$ in degrees?

Correct Answer: 50

Solution

The angles $$\angle BAC$$ and $$\angle BNC$$ stand on the same chord $$BC$$ with their vertices on the same side of it, so $$\angle BNC = \angle BAC = 80^\circ$$. Since $$N$$ lies on the perpendicular bisector of $$BC$$, we have $$NB = NC$$, so triangle $$BNC$$ is isosceles. Hence $$\angle NBC = \frac{180^\circ - 80^\circ}{2} = 50^\circ$$.

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