# Bisector of angle formed at the orthocentre passes through the circumcentre

BdMO 2012

In an acute angled triangle $ABC$, $\angle A= 60$. We have to prove that the bisector of one of the angles formed by the altitudes drawn from $B$ and $C$ passes through the center of the circumcircle of the triangle $ABC$

Here,I tried to draw the perpendicular bisectors of sides AC and AB and then note that there is a parallelogram is formed at the centre.If we can now prove that the parallelogram is a rhombus,we will be done.I am thinking about using the properties of $30-60-90$ triangle somehow.I also found four concyclic points in the figure.A hint will be appreciated.

All credits to Mick for the picture above.

• Sorry, cannot understand your statement. It would be helpful if a diagram is attached.
– Mick
May 12, 2014 at 3:41
• @Mick,assume that the perpendiculars from B and C intersect at O.Now we construct the bisectors of $\angle BOC$ and its supplementary angle.Then one of these lines passes through the circumcentre(or that's what we have to prove anyway) May 12, 2014 at 3:44

Let $P$ and $Q$ be, respectively, the orthocenter and circumcenter of $\triangle ABC$. Let $R$ be the intersection of the perpendicular bisector of $\overline{AB}$ with the altitude from $B$; and let $S$ be the intersection of the perpendicular bisector of ${AC}$ with the altitude from $C$.

Introduce $B^\prime = \overleftrightarrow{AB}\cap\overleftrightarrow{QS}$ and $C^\prime = \overleftrightarrow{AC}\cap\overleftrightarrow{QR}$. Then $\triangle ABC^\prime$ and $\triangle AB^\prime C$ are equilateral.

Observe that $R$ is orthocenter, circumcenter, and incenter for $\triangle ABC^\prime$; likewise, $S$ for $\triangle AB^\prime C$. Therefore, $R$ and $S$ lie on the bisector of $\angle A$. A little angle-chasing shows that $\triangle PRS$ and $\triangle QRS$ are equiangular, hence equilateral, so that $\square PSQR$ is a rhombus, and the result follows.

• +1, brilliant solution.Thanks a lot.May I know the motivation behind the constructions?Was your main intention to create an equilateral triangle? May 12, 2014 at 18:08
• @rah4927: I was playing with a dynamic GeoGebra sketch of the problem, and it occurred to me that $R$ and $S$ were on the bisector of $\angle A$. Having constructed the bisector to verify my suspicion, I found that I had a number of $30^\circ$-$60^\circ$-$90^\circ$ triangles at my disposal. Working with various proportions among those would get to the result, but as I started to think that through, I realized that $R$ and $S$ were the ortho-/circum-/in-centers of equilateral triangles, and then everything fell into place nicely. So ... No, I didn't think about the equilaterals right away.
– Blue
May 12, 2014 at 19:40
• @Blue Please explain why, through the introduction of B’ (and also C’), you can infer that △AB′C is also equilateral. I understand that S is its ortho-center and S lies on the angle bisector of ∠A.
– Mick
May 13, 2014 at 6:00
• @Mick: $B^\prime$ is on $\overleftrightarrow{QS}$, which is the perpendicular bisector of $\overline{AC}$. Thus, $B^\prime$ is equidistant from $A$ and $C$, making $\triangle AB^\prime C$ isosceles; since there's a $60^\circ$ angle at $A$, that triangle is in fact equilateral. (Likewise, $C^\prime$ is on the perpendicular bisector of $\overline{AB}$, etc, etc.)
– Blue
May 13, 2014 at 6:17
• @Blue I forgot that your diagram is different than mine. If I have the roles of B and C interchanged, then I need not ask the question. Thanx.
– Mick
May 13, 2014 at 17:14

I don't have the solution.

I am just trying to help by uploading the rough sketch according to the description in your comment.

• I have added your picture to the body of my question.I hope you don't mind(if you do,feel free to let me know about it and I will take it down) May 12, 2014 at 15:28
• Not at all. One thing to note is that the image was stretched such that some of the should-be right angles were not truly revealed. A typical example is the angles between the orange dotted lines.
– Mick
May 12, 2014 at 16:16

Here, $\angle BAC = 60^\circ$. So, $\angle BHC = \angle BA'C = 180^\circ - \angle A = 120^\circ = 2\times\angle A = \angle BOC$. So, $B,C,H,O$ are concyclic.

Thus, $\angle OHC=\angle OBC = 30^\circ$ .

Also, $\angle CHH_b = \angle CBH + \angle BCH = (90^\circ - \angle C) + (90^\circ - \angle B) = \angle A =60^\circ$

So, $\angle OHH_b = \angle OHC = 30^\circ$, proving $OH$ bisects $\angle CHH_b$.