Let ABC be a triangle with |AB| = |AC|. Let H and O be the
orthocenter and circumcenter of ΔABC respectively.

Prove that the circumcenter of ΔBHO lies on ray BA.
 

Let the diameter through A, O, H, cut BC at D and the circle again at E.
Let M and N be the mid-points of OH and AB respectively.
Let BH cut AC at F and let the perpendicular to AE at M cut AB at P.
Let triangle ABC have circum-radius r and let its angle at A be 2v.

Triangles ACD, BCF are similar so /CBF = v.
Also, /EBD = /EAC = v (angles in same segment).
Therefore /CBF = /EBD, triangles BDH and BDE are congruent, and
D is the mid-point of HE.
Since M is the mid-point of OH, it follows that MD = OE/2 = r/2.

MD is the projection of PB, so PB = MD/cos(v) = r/(2cos(v)).
Thus: OB/PB = r/PB = 2cos(v).
But AB/OB = 2cos(v), hence OB/PB = AB/OB, and it follows that
triangles PBO and OBA are similar (SAS) since /PBO is common.

Using this similarity, since AO = BO, it now follows that OP = BP.
Also, HP = OP (by symmetry about MP), thus OP = BP = HP and the
point P, which lies on AB, must be the circumcentre of triangle BHO.

Posted by Harry on 2013-10-16 12:07:50