Orthocenter Reflection over a Midpoint

A proof that reflecting the orthocenter H over the midpoint of a side places the reflection on the circumcircle, forming a parallelogram.

Keywords: orthocenter, reflection, midpoint, parallelogram, circumcircle

Prerequisites: circumcenter-proof, cyclic-quadrilateral-proof, bhc-identity-proof · Difficulty: advanced

Reflecting the orthocenter $H$ over the midpoint $M_A$ of side $BC$ gives a point $H^{\prime }$ that lies on the circumcircle.

Strategy: since $M_A$ is the midpoint of both $BC$ and $H{H}^{\prime }$, the quadrilateral $B{H}^{\prime }CH$ is a parallelogram (its diagonals bisect each other). Then $B{H}^{\prime }\parallel CH$ and $C{H}^{\prime }\parallel BH$; combined with the altitudes $BH\perp AC$ and $CH\perp AB$, this places $H^{\prime }$ on the circumcircle, diametrically opposite $A$.

Step 1: Setup

Start with triangle $ABC$, its circumcircle, and the orthocenter $H$. Mark the midpoint $M_A$ of side $BC$.

Step 2: Reflect $H$ over $M_A$

Reflect $H$ over the midpoint $M_A$ to get $H^{\prime }$. Since $M_A$ is the midpoint of $H{H}^{\prime }$, and also the midpoint of $BC$, the diagonals of quadrilateral $B{H}^{\prime }CH$ bisect each other.

Step 3: $H^{\prime }$ on the Circumcircle

Claim: The reflection $H^{\prime }$ of $H$ over the midpoint $M_A$ of $BC$ lies on the circumcircle of $△ABC$, diametrically opposite $A$.

Proof strategy. Since $M_A$ is the midpoint of both $H{H}^{\prime }$ and $BC$, the diagonals of $B{H}^{\prime }CH$ bisect each other, so $B{H}^{\prime }CH$ is a parallelogram. From the parallel sides we get $C{H}^{\prime }\perp CA$ and $B{H}^{\prime }\perp AB$; the converse of the cyclic quadrilateral theorem then places $H^{\prime }$ on the circumcircle.

Step 4: Parallelogram $B{H}^{\prime }CH$

Since the diagonals $BC$ and $H{H}^{\prime }$ bisect each other at $M_A$, the quadrilateral $B{H}^{\prime }CH$ is a parallelogram. This means:

• $B{H}^{\prime }\parallel CH$ and $B{H}^{\prime }=CH$
• $C{H}^{\prime }\parallel BH$ and $C{H}^{\prime }=BH$

Step 5: $H^{\prime }$ on the Circumcircle

Since $BH$ is part of the altitude from $B$ (perpendicular to $CA$), and $C{H}^{\prime }\parallel BH$, we get $C{H}^{\prime }\perp CA$, making $\angle AC{H}^{\prime }=90°$.

Similarly, $CH$ is part of the altitude from $C$ (perpendicular to $AB$), and $B{H}^{\prime }\parallel CH$, we get $B{H}^{\prime }\perp AB$, making $\angle AB{H}^{\prime }=90°$.

In quadrilateral $AB{H}^{\prime }C$, the opposite angles at $B$ and $C$ now sum to $180°$. By the converse of the cyclic quadrilateral theorem, $AB{H}^{\prime }C$ is cyclic — $H^{\prime }$ lies on the circumcircle of $△ABC$, diametrically opposite $A$.

Reflecting $H$ over the midpoint of any side gives a point on the circumcircle. The key insight is the parallelogram formed by $H$, the two adjacent vertices, and the reflection.

This property is the foundation of the nine-point circle proof.

Alternative proof

Using the orthocenter angle identity ($\angle BHC=180°-\angle A$), the parallelogram $B{H}^{\prime }CH$ gives a shorter route to the same conclusion:

1. Opposite angles of a parallelogram are equal, so $\angle B{H}^{\prime }C=\angle BHC$.
2. By the orthocenter angle identity, $\angle BHC=180°-\angle A$, hence $\angle B{H}^{\prime }C=180°-\angle A$.
3. Then $\angle BAC+\angle B{H}^{\prime }C=\angle A+(180°-\angle A)=180°$, so by the converse of the cyclic quadrilateral theorem, $AB{H}^{\prime }C$ is cyclic — $H^{\prime }$ lies on the circumcircle.

This is the same chain used in the side-reflection proof, with the parallelogram supplying the angle equality $\angle B{H}^{\prime }C=\angle BHC$ instead of reflection symmetry. ∎