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)
Drawing: https://pasteboard.co/KbDzeCx.jpg
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.
Best Answer
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 $\triangle BOC$
$$\theta=30$$