Rmc, 2020

Geometry Difficulty 6.1 National Olympiad Prove it Romania

Let $D$ be a point in the interior of triangle $ABC$ such that $\angle BAD = 40^\circ$ , $\angle DAC = 30^\circ$ , $\angle BCD = 20^\circ$ and $\angle DCA = 50^\circ$ . Find $\angle CBD$ .

Solution

Observe that $ABC$ is an isosceles triangle, with $BA = BC$ and $\angle ABC = 40^\circ$ . Suppose that the perpendicular bisector of the triangle's base intersects $AD$ at $T$ .
Since $\angle DAC < \angle DCA$ , we have $CD < DA$ therefore $T \in (AD)$ and $\angle TBC = 20^\circ$ .

We will prove that $D$ is the incenter of triangle $BTC$ , hence $BD$ bisects $\angle TBC$ , and it follows that $\angle CBD = 10^\circ$ .

Because $TCA$ is an isosceles triangle, we have $\angle TCA = 30^\circ$ , and hence $\angle TCD = 20^\circ = \angle BCD$ . It follows that $CD$ bisects $\angle BCT$ .

A short computation shows that $\angle CTB = \angle BTA = 120^\circ$ and $\angle DTC = 60^\circ$ , therefore $TD$ is the angle bisector of $\angle BTC$ .

We conclude that $D$ is the incenter of triangle $BCT$ , as desired.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.