Prove that the symmetric point of orthocenter $H$ of a triangle with respect to the midpoint of any side resides on circumcircle.

Question

Prove that the symmetric point of orthocenter $H$ of a triangle with respect to the midpoint of any side resides on the triangle's circumcircle.

In the above figure,it's sufficient to prove that triangle $BHM$ is equal to triangle $CKM$, but how? I tried to make use of the symmetric point of $H$ with respect to $BC$ which resides on circumcircle but failed...

Jack D'Aurizio · Accepted Answer · 2016-10-15T16:26:21.487

3

The symmetric of $H$ with respect to the $BC$-side, say $J$, lies on the circumcircle of $ABC$ since $\widehat{BHC}+\widehat{BAC}=\pi$. If we consider the symmetric of $J$ with respect to the perpendicular bisector of $BC$ we get $K$, hence $K$ lies on the circumcircle of $ABC$, too, since the circumcircle of $ABC$ is symmetric with respect to the perpendicular bisector of $BC$.

As an alternative, from the statement it clearly follows that $BHCK$ is a parallelogram, since its diagonals $BC$ and $HK$ intersect at their midpoints. That gives $\widehat{CKB}=\widehat{BHC}=\pi-\widehat{BAC}$, hence $K$ lies on the circumcircle of $ABC$.

edited Oct 15 '16 at 16:26

answered Oct 15 '16 at 14:08

Jack D'Aurizio

353,855

Please explain more – Hamid Reza Ebrahimi Oct 15 '16 at 16:02
@HamidRezaEbrahimi: what is not clear? $\widehat{BHC}=\pi-\widehat{BAC}$ follows from angle chasing, and it gives that $BACJ$ is a cyclic quadrilateral, from which $J\in\Gamma$. – Jack D'Aurizio Oct 15 '16 at 16:06
Why $K$ lies on circumcircle?! I don't understand it's reason! – Hamid Reza Ebrahimi Oct 15 '16 at 16:11
1

@HamidRezaEbrahimi: $J$ lies on the circumcircle. $K$ is the symmetric of $J$ with respect to the perpendicular bisector of $BC$. The circumcircle is symmetric with respect to such a perpendicular bisector. $$J\in\Gamma\quad \Longrightarrow\quad K\in\Gamma$$ – Jack D'Aurizio Oct 15 '16 at 16:13

score 1 · Answer 2 · 2016-10-15T15:15:51.437

1

Use that BHCK is a quadrilateral whose diagonals intersect in the middle. So is a parallelogram. Then show that BHCK is inscribed.

edited Oct 15 '16 at 15:15

answered Oct 15 '16 at 13:00

We are to prove $HM=MK$... – Hamid Reza Ebrahimi Oct 15 '16 at 13:04
@WebersonArcanjo: Edit your answer to include your own comment, otherwise I fear that your post will get deleted. – Alex M. Oct 15 '16 at 14:40

score 1 · Answer 3 · answered Aug 17 '19 at 08:56

A complex number proof. Without loss of generality we may assume that $A,B,C$ are complex numbers with the unit circle of $\mathbb{C}$ centered at the origin $O$. Then $K=-C$, $H=A+B+C$ and $$\frac{H+K}{2}=\frac{A+B}{2}=M.$$ As a reference I suggest Bashing Geometry with Complex Numbers by E. Chen.

Prove that the symmetric point of orthocenter $H$ of a triangle with respect to the midpoint of any side resides on circumcircle.

3 Answers3