 Given two adjacent angles, AOB and BOC, each of 60 degrees, construct the bisector of angle BCO, which intersects OA in D. Find the angle BDC.

Source (Romanian Math Magazine - Gazeta Matematica; online competition)

Trying the drawing in Geogebra, I found the angle is 30 degrees, regardless of value of angle C (moving B on the bisector OB). Is there some theorem which I could use to derive this?

From the figure, BDO = 90 - BCD (also don't know how to prove it).

I am interested in a solution for elementary school students. Theorem : In a triangle bisector of an interior angle and bisector of other two exterior angles meet at a point.

$$\angle AOM=60$$

Since $$CD$$ is the interior bisector of $$\angle BOC$$ and $$OD$$ is the exterior bisector of $$\angle BOD$$ of $$\triangle BOC$$ (one interior bisector and one exterior bisector) meet at $$D$$, $$BD$$ is the bisector of $$\angle OBN$$.

$$\alpha +60=\beta + \theta$$

$$2\alpha +60=2\beta,\alpha +30=\beta (*)$$ $$(*)$$exterior $$\angle$$ of $$\triangle BOC$$

$$\theta=30$$

Added: As @lionheart pointed out, wrt $$\triangle OBC$$, the internal angle bisector of $$\angle OCB$$, and the external angle bisector of $$\angle COB$$ meet at D, then DB is the external angle bisector of $$\angle CBO$$. Thus, we have $$s = s$$ as shown. After producing DB to cut the y-axis at T, we can always draw the circle passing through T, D, O. Next, we need to prove that C is also a con-cyclic point of that circle. This is done by the following:-

Using "exterior angle of triangle" twice, we have $$s = 30^0 + t$$ and $$2s = 2x + 60^0$$. Consequently, $$x = t$$.

On one hand, we have $$x + y = 60^0$$

On the other hand, we have $$t + 90^0 + y + z = 180^0$$

Result follows.