Here is a geometry problem posted on MathematicsArt and the Facebook feed Art of Mathematics.
Let $ABC$ be a triangle. Let $P$ be on $AB$ so that $CP\perp AB$. Let $D$ be on $CP$ so that $\angle PAD=\angle DAC=10^\circ$. $\angle PCB=40^\circ$. Required to find $\angle DBP=x$.
Angle-chasing gives $\angle CBP=50^\circ$.
The solution just derives the trig equation $\tan 20^\circ \tan x = \tan 10^\circ \tan 50^\circ$ and numerically calculates $x$ as $30^\circ$. No doubt there is an algebraic expression for the values of each of the various tangents involved, from which the answer could be derived. But my question does not concern such a derivation.
My question is: how, if at all, can the problem be solved using only elementary geometry? (Euclid, Elements, books 1 and 3).
The fact that $\angle DBP=30^\circ$ suggests constructing $E$ on $DP$ produced, so that $AB$ bisects $DE$, then proving that $\triangle BDE$ is equilateral, but I don't see how to prove this. Alternatively, either construct an equilateral $\triangle DBG$ and prove $BG||PC$, or construct $BH||PC$ and $H$ on this parallel so that $BD=BH$, then prove that $\triangle BDH$ is equilateral, but again I don't see how to prove this.
Let $E$ be the reflection of $D$ in $AB$. Let $O$ be the circumcenter of $\triangle ACE$. Note that $\angle CEA = \angle ADE = 80^\circ$, hence $\angle OAC = 90^\circ - \angle CEA = 10^\circ = \angle DAC$. This shows that $O$ lies on $AD$.
Moreover, $\angle EAC = 30^\circ$, hence $\angle EOC = 2\angle EAC = 60^\circ$. Since $OE=OC$, triangle $EOC$ is equilateral.
Let $OE$ intersect $AB$ at $F$. We have $\angle FDE = \angle DEF = 60^\circ$, hence triangle $DEF$ is equilateral. Now we see that $CDFO$ is an isosceles trapezoid. Therefore $\angle FCD = \angle FOD = 2\angle EAO = 40^\circ=\angle PCB$. This shows that $CF$ is symmetric to $CB$ with respect to $CE$. In particular, $F$ is symmetric to $B$. Therefore triangle $DBE$ is equilateral, hence $\angle DBP = \frac 12 \cdot \angle DBE = 30^\circ$.