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.

Proof Strategy

Reflect $H$ over side $BC$ to get $H'$. We show $\angle BH'C = 180° - \angle A$. Since the inscribed angle subtending arc $BC$ (not containing $A$) from the circumcircle also equals $180° - \angle A$, the point $H'$ must lie on the circumcircle.

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

The Proof

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 $H'$. The midpoint of $HH'$ lies on $BC$.

Step 3: $H'$ on the Circumcircle

Claim: The reflection $H'$ of $H$ over side $BC$ lies on the circumcircle of $\triangle ABC$.

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

Step 4: Calculate $\angle BH'C$

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

Reflection over $BC$ preserves angles, so $\angle BH'C = \angle BHC = 180° - \angle A$.

Step 5: $H'$ 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 $\angle BH'C = 180° - \angle A$ and $H'$ is on the correct side of $BC$, the point $H'$ 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.

Conclusion

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

This connects the orthocenter intimately to the circumcircle.