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.
Geometry question in HK IMO prelim 2018
anglecontest-mathgeometrytriangles
Related Solutions
Observe Triangles $ACB$ and $ADC$.
$\angle ACB$ is common. And $AB=CD$, $AC=AC$(Common).
Therefore the triangles are similar (Two sides have lengths in the same ratio, and the angles included between these sides have the same measure).
Which means,
$\frac{AB}{AC}=\frac{AC}{AB}$, and the angle between them is common( $\angle ACB$)
Now,
$\angle CAD = \angle ACB =c$
Thus, $a=c$
Therefore,
$5a=180^0$. $a=36^0$ and $\angle BAC=72^0$.
Let $D$ be placed on the line $BC$ such that $B$ is placed between $D$ and $C$.
Thus, since $$\measuredangle DBA=42^{\circ}=\measuredangle ABE,$$ we see that $BA$ is a bisector of $\angle DBE$ and since $EF$ is a bisector of $\angle AEB$, we obtain that
$CF$ is a bisector of $\angle BCA.$
Id est, $$\measuredangle BFC=12^{\circ}+18^{\circ}=30^{\circ}.$$
Best Answer
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 $$\measuredangle BEC=12^{\circ}+18^{\circ}=30^{\circ}.$$