The Orthocenter Angle Identity

A proof that in any triangle, the angle at the orthocenter subtended by two vertices equals 180° minus the angle at the third vertex.

Keywords: orthocenter, angle identity, altitude, quadrilateral angle sum, vertical angles

Prerequisites: orthocenter-proof · Difficulty: intermediate

Theorem: If $H$ is the orthocenter of $\triangle ABC$, then $$\angle BHC = 180° - \angle A.$$

Proof Strategy

Let $F_b$ be the foot of the altitude from $B$ on $CA$, and $F_c$ the foot of the altitude from $C$ on $AB$. Consider the quadrilateral $AF_cHF_b$:

  1. Two of its interior angles are right angles (altitudes are perpendicular to the opposite sides): $\angle AF_cH = \angle AF_bH = 90°$.
  2. A third angle is $\angle F_cAF_b = \angle A$.
  3. Interior angles of a quadrilateral sum to $360°$, so $\angle F_cHF_b = 360° - 90° - 90° - \angle A = 180° - \angle A$.
  4. The points $B$, $H$, $F_b$ are collinear (they lie on the altitude from $B$) and so are $C$, $H$, $F_c$. So $\angle BHC$ and $\angle F_cHF_b$ are vertical angles at $H$, hence equal.

The same argument applied at $B$ and $C$ gives $\angle AHC = 180° - \angle B$ and $\angle AHB = 180° - \angle C$.

The Proof

Step 1: Triangle with Orthocenter

Start with triangle $ABC$. Drop the altitudes from $B$ and $C$, and mark their intersection as the orthocenter $H$.

Step 2: $\angle BHC = 180° - \angle A$

For any triangle $ABC$ with orthocenter $H$: $$\angle BHC = 180° - \angle A.$$

By the same argument at the other vertices, $\angle AHC = 180° - \angle B$ and $\angle AHB = 180° - \angle C$.

Proof sketch. Look at quadrilateral $AF_cHF_b$: two of its angles are $90°$ (altitudes), one is $\angle BAC$, and the four interior angles of any quadrilateral sum to $360°$, so $\angle F_cHF_b = 180° - \angle A$. Since $\angle BHC$ is vertical to $\angle F_cHF_b$ at $H$, the two are equal.

Step 3: Quadrilateral $AF_cHF_b$

Look at the quadrilateral formed by $A$, $F_c$, $H$, $F_b$. Three of its four interior angles are already known.

Step 4: The Fourth Angle

Interior angles of a quadrilateral sum to $360°$. Subtracting the two right angles and $\angle A$: $$\angle F_cHF_b = 360° - 90° - 90° - \angle A = 180° - \angle A.$$

Step 5: Vertical Angles at $H$

$B$, $H$, $F_b$ lie on the altitude from $B$, and $C$, $H$, $F_c$ on the altitude from $C$. So at $H$, the rays $HB$ / $HF_b$ are opposite, and so are $HC$ / $HF_c$. That makes $\angle BHC$ and $\angle F_cHF_b$ vertical angles, hence equal.

Step 6: $\angle BHC = 180° - \angle A$

Vertical angles are equal, so $\angle BHC = \angle F_cHF_b = 180° - \angle A$.

By the same argument at $B$ and $C$: $\angle AHC = 180° - \angle B$ and $\angle AHB = 180° - \angle C$.

Conclusion

For any triangle $ABC$ with orthocenter $H$: $$\angle BHC = 180° - \angle A.$$

The proof uses only two elementary facts — altitudes are perpendicular to opposite sides, and interior angles of a quadrilateral sum to $360°$ — plus the vertical-angle relation at $H$.

This identity is a key lemma: it explains why reflecting $H$ over a side lands on the circumcircle, and it appears throughout orthocenter / nine-point-circle geometry.