# What are the possible values of $\angle CAB$?

In an acute $\triangle ABC$, the segment $CD$ is an altitude and H is the orthocentre. Given that the circumcentre of the triangle lies on the line containing the bisector of the $\angle DHB$, determine all possible values of $\angle CAB$ .

Observe that the exterior bisector of $\angle BHC$ meets the perpendicular bisector of $BC$ in the midpoint of the arc $BHC$ of circle $BHC$ so $BHOC$ is concyclic [$O$ is the circumcentre] so $180-\angle BAC = 2\times \angle BAC$ so $\angle BAC=60^{\circ}$.

• You got it very early. +1 Dec 11, 2016 at 13:02
• @THELONEWOLF. It was very easy.
– user371838
Dec 11, 2016 at 13:03
• I know. I was just writing the answer, but now there is already one so, no need . Dec 11, 2016 at 13:04
• Could there be some typo in the statement $\angle BOH = \angle BOC$.
– Mick
Dec 11, 2016 at 16:35
• Please mark the part as edited instead of correcting it directly. Otherwise, others will think I am saying something nonsense.
– Mick
Dec 11, 2016 at 17:11