# Triangle and point inside

I need elementary (no use of trigonometry is allowed) solution to the following problem.

Given a triangle ABC with $\angle BAC = 30$ and $\angle ABC=40$. $P$ is inside the triangle and $\angle PAB=\angle PBA=10^{\circ}$. Find $\angle PCB$.

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I don't think that you can avoid trigonometry in general for this kind of question, so you would have to use some special feature of the exact numerical values (also, it is not hard to draw a picture and determine the angle which helps in finding a way to prove it). – Phira May 23 '12 at 23:00
I've deleted my partial answer; I'll put it back if I get around to finishing it. – Michael Hardy May 24 '12 at 17:46

Prolong the side AC after C until a point X with the property that $\angle CXB = 80^\circ$.

Now, you can easily calculate $\angle ABX = 70^\circ$ (angle sum in ABX) and thus $$\angle PBX = 60^\circ.$$

But you also know that $\angle APB = 160^\circ$ and $\angle AXB=80^\circ$. This implies that $P$ is the center of the circumcircle of the triangle ABX and this implies that $\angle BPX=2 \angle BAX= 60^\circ$ (I use the theorem on inscribed angles in both directions here). So we get

$$\angle BPX = 60^\circ.$$

Therefore, the triangle BPX is equilateral and BC (making angles of $30^\circ$) is an axis of symmetry for the quadrangle PBXC.

Therefore, the angle $\angle BCP = \angle BCX=70^\circ$ (angle sum in BCX) which is the desired value.

Note that this proof does not really avoid trigonometry. The idea is that for particular values, you can let an equilateral triangle do the trigonometric work where it is trivial. Note also that I first checked the size $70^\circ$ on geogebra and then decided on adding the point X by a reflection and then wrote down the proof in reverse order of finding it.

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I had begun to think about the circumcircle of $\triangle ABC$, and if I'm not mistaken, this line $BX$ is tangent to that circle. Dunno if that gets us anywhere. But I've spent too much time on this. – Michael Hardy May 24 '12 at 0:00