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 · Difficulty: advanced

Theorem: Reflecting the orthocenter $H$ over the midpoint $M_A$ of side $BC$ gives a point $H'$ that lies on the circumcircle.

Proof Strategy

Since $M_A$ is the midpoint of both $BC$ and $HH'$, the quadrilateral $BH'CH$ is a parallelogram (diagonals bisect each other).

This means $BH' \parallel CH$ and $CH' \parallel BH$. Since $BH \perp AC$ (altitude), we get $CH' \perp AC$. Similarly $BH' \perp AB$. This places $H'$ on the circumcircle diametrically opposite $A$.

This construction is foundational for the nine-point circle theorem.

The Proof

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 $H'$. Since $M_A$ is the midpoint of $HH'$, and also the midpoint of $BC$, the diagonals of quadrilateral $BH'CH$ bisect each other.

Step 3: $H'$ on the Circumcircle

Claim: The reflection $H'$ of $H$ over the midpoint $M_A$ of $BC$ lies on the circumcircle of $\triangle ABC$, diametrically opposite $A$.

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

Step 4: Parallelogram $BH'CH$

Since the diagonals $BC$ and $HH'$ bisect each other at $M_A$, the quadrilateral $BH'CH$ is a parallelogram. This means:

Step 5: $H'$ on the Circumcircle

Since $BH$ is part of the altitude from $B$ (perpendicular to $CA$), and $CH' \parallel BH$, we get $CH' \perp CA$, making $\angle ACH' = 90°$.

Similarly, $CH$ is part of the altitude from $C$ (perpendicular to $AB$), and $BH' \parallel CH$, we get $BH' \perp AB$, making $\angle ABH' = 90°$.

In quadrilateral $ABH'C$, the opposite angles at $B$ and $C$ now sum to $180°$. By the converse of the cyclic quadrilateral theorem, $ABH'C$ is cyclic — $H'$ lies on the circumcircle of $\triangle ABC$, diametrically opposite $A$.

Conclusion

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 BHC identity ($\angle BHC = 180° - \angle A$), the parallelogram $BH'CH$ gives a shorter route to the same conclusion:

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

This is the same chain used in the side-reflection proof, with the parallelogram supplying the angle equality $\angle BH'C = \angle BHC$ instead of reflection symmetry.