In triangle ABC, $\angle CAB = 18$ and $ \angle BCA = 24$ .$ E $ is a point on $CA $ such that $\angle CEB=60$ and $F $is a point on $AB $ such that $\angle AEF=60$.What is the
measure, in degrees, of $\angle BFC$?
Taken from the 2014 IMC
Ive solved similar problems but im not sure how to solve this
I worked out $\angle BAC= 18$ , $\angle EBC =96$ , $\angle FEB = 42$, and $\angle BEC = \angle FEB = \angle AEF=60$.
Best Answer
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, $$\measuredangle BFC=12^{\circ}+18^{\circ}=30^{\circ}.$$