The Concurrency of the Altitudes in a Triangle
A Trigonometric Proof

Dušan Vallo
February 2012

Let \(ABC\) be a triangle. Using the standard notations, we denote the altitudes \(h_a = AH_a\), \(h_b = BH_b\), \(h_c = CH_c\).


The three altitudes of a triangle are concurrent.


First observe that in a right-angled triangle ABC the altitudes do meet at the right angled vertex. Now, suppose that triangle ABC is not right-angled and denote \(D = h_a \cap h_b\), \(E = h_a\cap h_c\). We wish to show that \(D\) coincides with \(E\).

illustration for a trigonometric proof of the concurrence of the altitudes in a triangle

The triangles \(ABH_a\), \(ACH_a\) are right-angled and we easy derive


\(h_a = c \cdot \sin B = b \cdot \sin C\).

From the right-angled triangles \(ACH_a\), \(ABH_b\) we obtain


\(CH_a = b \cdot \cos C,\) \(AH_b = c \cdot \cos A.\)

The triangles \(CEH_a\), \(ADH_b\) are also right-angled and so \(\angle ADH_b = \angle C\), \(\angle CEH_a = \angle B\). From (2), we obtain

\(\displaystyle AD=\frac{c\cdot \cos A}{\sin C}\), \(\displaystyle EH_a=\frac{b\cdot \cos C}{\tan B}\).

Let \(x=DE\). Then, for the altitude \(AH_a\),

\(AH_a = AD + x + EH_a\).

Using (1), this can be rewritten as


\(\displaystyle c\cdot\sin C=c\cdot\frac{\cos A}{\sin C}+x+b\cdot\frac{\cos C}{\tan B}\).

Next we apply the theorem about the sum of angles in a triangle and well-known formulas

\(\cos (X+Y)=\cos X\cdot\cos Y-\sin X\cdot\sin Y\),
\(\cos (\pi -X)=-\cos X.\)

We use the latter in (3) first with \(X = A=\pi - (B+C)\) to eventually arrive at

\(\displaystyle x=(b\cdot\cos C-c\cdot\cos B)\bigg[(1 - \frac{1}{\tan B\cdot\tan C}\bigg]\).

By (1), the first factor equals \(b\cdot\cos C-c\cdot\cos B=h_a - h_a=0\), implying \(x=0\), as required. This completes the proof.


|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny