Orthocenter Reflection over a Side

A proof that reflecting the orthocenter H over a side of the triangle places the reflection on the circumcircle.

Keywords: orthocenter, reflection, circumcircle, inscribed angle

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

Theorem: the reflection of the orthocenter $H$ over any side of the triangle lies on the circumcircle.

Strategy: reflect $H$ over side $BC$ to get $H^{\prime }$ and compute $\angle B{H}^{\prime }C=180°-\angle A$. The inscribed angle subtending arc $BC$ (not containing $A$) also equals $180°-\angle A$, so $H^{\prime }$ must lie on the circumcircle.

Step 1: The Circumcircle and Orthocenter

Start with triangle $ABC$, its circumcircle centered at $O$, and its orthocenter $H$.

Step 2: Reflect $H$ over $BC$

Reflect the orthocenter $H$ over side $BC$ to obtain point $H^{\prime }$. The midpoint of $H{H}^{\prime }$ lies on $BC$.

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

Claim: The reflection $H^{\prime }$ of $H$ over side $BC$ lies on the circumcircle of $△ABC$.

Proof strategy. We compute $\angle B{H}^{\prime }C$ and find it equals $180°-\angle A$, using the orthocenter angle identity ($\angle BHC=180°-\angle A$) plus the fact that reflection preserves angles. The inscribed-angle theorem then forces $H^{\prime }$ onto the arc $BC$ opposite $A$.

Step 4: Calculate $\angle B{H}^{\prime }C$

The orthocenter angle identity (proven separately) states $\angle BHC=180°-\angle A$ in any triangle.

Reflection over $BC$ preserves angles, so $\angle B{H}^{\prime }C=\angle BHC=180°-\angle A$.

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

By the inscribed angle theorem, a point $P$ on the circumcircle on the opposite side of $BC$ from $A$ satisfies $\angle BPC=180°-\angle A$.

Since $\angle B{H}^{\prime }C=180°-\angle A$ and $H^{\prime }$ is on the correct side of $BC$, the point $H^{\prime }$ lies on the circumcircle!

Step 6: Verify for Other Sides

By the same argument, reflecting $H$ over sides $CA$ and $AB$ also places the reflection on the circumcircle. This property connects the orthocenter to the circumcircle in a fundamental way.

The reflection of $H$ over any side lies on the circumcircle:

• $H_a^{\prime }$ (reflection over $BC$) lies on the circumcircle
• $H_b^{\prime }$ (reflection over $CA$) lies on the circumcircle
• $H_c^{\prime }$ (reflection over $AB$) lies on the circumcircle

This connects the orthocenter intimately to the circumcircle. ∎

Notes

This is a key lemma used in proofs about the nine-point circle and Euler line.