Find angle in quadrilateral

Question

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.

Answer 1

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 $\triangle BOC$

$$\theta=30$$

Answer 2

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, $x = t$.

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

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

Result follows.