geometry question in HK IMO prelim 2018

Question

2018 IMO prelim in HK, Q.3: In triangle ABC, ∠ BAC = 18° and angle ∠BCA = 24°. D is a point on AC such that ∠BDC = 60°. If the bisector of ∠ADB meets AB at E, find ∠BEC.

Any form of help will be appreciated.

Answer 1

Also, there is a nice geometric solution.

Indeed, let $F\in BC$ such that $B$ placed between $C$ and $F$.

Thus, since $$\measuredangle DBA=\measuredangle ABF=42^{\circ}$$ and $$\measuredangle BDE=\measuredangle ADE=60^{\circ},$$ we obtain that $CE$ is a bisector of $\angle ACB$, which gives $\measuredangle ACE=12^{\circ},$ which says $$\measuredangle BEC=12^{\circ}+18^{\circ}=30^{\circ}.$$

Answer 2

Let $BD\cap CE=\{P\}$ and $\measuredangle BEC=x$.

Thus, $$\measuredangle CDB=\measuredangle BDE=\measuredangle ADE=60^{\circ},$$ which says that $DP$ is a bisector of $\angle CDE$.

Also, easy to see that $\measuredangle DCE=x-18^{\circ},$ $\measuredangle BCE=42^{\circ}-x,$ $\measuredangle CBP=96^{\circ},$ $\measuredangle PBE=42^{\circ}$ and $\measuredangle PED=78^{\circ}-x.$

Id est, by law of sines we obtain: $$\frac{\sin(x-18^{\circ})}{\sin(78^{\circ}-x)}=\frac{ED}{CD}=\frac{EP}{PC}=\frac{\frac{EP}{BP}}{\frac{PC}{BP}}=\frac{\frac{\sin42^{\circ}}{\sin{x}}}{\frac{\sin96^{\circ}}{\sin(42^{\circ}-x)}},$$ which gives $$\sin96^{\circ}\sin(x-18^{\circ})\sin{x}=\sin42^{\circ}\sin(42^{\circ}-x)\sin(78^{\circ}-x)$$ or $$2\sin48^{\circ}\left(\cos18^{\circ}-\cos(2x-18^{\circ})\right)=\cos36^{\circ}-\cos\left(120^{\circ}-2x\right)$$ or $$\sin66^{\circ}+\sin30^{\circ}-\sin(30^{\circ}+2x)-\sin(66^{\circ}-2x)=\cos36^{\circ}+\cos\left(60^{\circ}+2x\right)$$ or $$\cos24^{\circ}-\cos36^{\circ}+\sin30^{\circ}-\sin(66^{\circ}-2x)-\cos\left(60^{\circ}-2x\right)-\cos\left(60^{\circ}+2x\right)=0$$ or $$\sin6^{\circ}+\sin30^{\circ}-\sin(66^{\circ}-2x)-\cos2x=0$$

$$\sin6^{\circ}-\sin(66^{\circ}-2x)+\cos60^{\circ}-\cos2x=0$$ or $$\sin(x-30^{\circ})\cos(36^{\circ}-x)+\sin(x-30^{\circ})\sin(x+30^{\circ})=0$$ or $$\sin(x-30^{\circ})(\cos(36^{\circ}-x)+\sin(x+30^{\circ}))=0$$ and since $18^{\circ}<x<42^{\circ},$ we obtain $$\sin(x-30^{\circ})=0$$ or $$x=30^{\circ}.$$