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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 $\angle CAB$ .

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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 $\angle BAC=60^{\circ}$.

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

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