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 \(h_a = AH_a\), \(h_b = BH_b\), \(h_c = CH_c\).
Theorem
The three altitudes of a triangle are concurrent.
Proof
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 \(E\).
The triangles \(ABH_a\), \(ACH_a\) are right-angled and we easy derive
From the right-angled triangles \(ACH_a\), \(ABH_b\) we obtain
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
\(AD=\frac{c\cdot \text{cos}A}{\text{sin}C}\), \(EH_a=\frac{b\cdot \text{cos}C}{\text{tan}B}\).
Let \(x=DE\). Then, for the altitude \(AH_a\),
\(AH_a = AD + x + EH_a\).
Using (1), this can be rewritten as
Next we apply the theorem about the sum of angles in a triangle and well-known formulas
\(\text{cos}(X+Y)=\text{cos}X\cdot\text{cos}Y-\text{sin}X\cdot\text{sin}Y\),
\(\text{cos}(\pi -X)=-\text{cos}X.\)
We use the latter in (3) first with \(X = A=\pi - (B+C)\) to eventually arrive at
\(x=(b\cdot\text{cos}C-c\cdot\text{cos}B)\big[(1 - \frac{1}{\text{tan}B\cdot\text{tan}C}\big]\).
By (1), the first factor equals \(b\cdot\text{cos}C-c\cdot\text{cos}B=h_a - h_a=0\), implying \(x=0\), as required. This completes the proof.
Trigonometry
- What Is Trigonometry?
- Addition and Subtraction Formulas for Sine and Cosine
- The Law of Cosines (Cosine Rule)
- Cosine of 36 degrees
- Sine and Cosine of 15 Degrees Angle
- Sine, Cosine, and Ptolemy's Theorem
- arctan(1) + arctan(2) + arctan(3) = π
- arctan(1/2) + arctan(1/3) = arctan(1)
- Morley's Miracle
- Napoleon's Theorem
- A Trigonometric Solution to a Difficult Sangaku Problem
- Trigonometric Form of Complex Numbers
- Derivatives of Sine and Cosine
- ΔABC is right iff sin²A + sin²B + sin²C = 2
- Advanced Identities
- Hunting Right Angles
- Point on Bisector in Right Angle
- Trigonometric Identities with Arctangents
- The Concurrency of the Altitudes in a Triangle - Trigonometric Proof
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