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

If $H$ is the orthocenter of $△ABC$, the angle it subtends at two vertices relates to the angle at the third:

Strategy: look at quadrilateral $A{F}_{c}H{F}_b$ formed by $A$ and the two altitude feet $F_b$, $F_c$. Two of its angles are right angles and a third is $\angle A$, so the fourth is $180°-\angle A$. Since $\angle BHC$ is vertical to that fourth angle at $H$, the two are equal.

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 $A{F}_{c}H{F}_b$: two of its angles are $90°$ (altitudes), one is $\angle BAC$, and the four interior angles of any quadrilateral sum to $360°$, so $\angle F_{c}H{F}_b=180°-\angle A$. Since $\angle BHC$ is vertical to $\angle F_{c}H{F}_b$ at $H$, the two are equal.

Step 3: Quadrilateral $A{F}_{c}H{F}_b$

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 $\angle A$: $$\angle F_{c}H{F}_b=360°-90°-90°-\angle A=180°-\angle A.$$

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 $HB$ / $H{F}_b$ are opposite, and so are $HC$ / $H{F}_c$. That makes $\angle BHC$ and $\angle F_{c}H{F}_b$ vertical angles, hence equal.

Step 6: $\angle BHC=180°-\angle A$

Vertical angles are equal, so $\angle BHC=\angle F_{c}H{F}_b=180°-\angle A$.

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

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. ∎