# 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

(1)

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

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

(2)

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

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

(3)

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

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.

### Trigonometry

- What Is Trigonometry?
- Addition and Subtraction Formulas for Sine and Cosine
- The Law of Cosines (Cosine Rule)
- Cosine of 36 degrees
- Tangent of 22.5
^{o}- Proof Wthout Words - Sine and Cosine of 15 Degrees Angle
- Sine, Cosine, and Ptolemy's Theorem
- arctan(1) + arctan(2) + arctan(3) = π
- Trigonometry by Watching
- 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
- Butterfly Trigonometry
- Binet's Formula with Cosines
- Another Face and Proof of a Trigonometric Identity
- cos/sin inequality
- On the Intersection of kx and |sin(x)|
- Cevians And Semicircles
- Double and Half Angle Formulas
- A Nice Trig Formula
- Another Golden Ratio in Semicircle
- Leo Giugiuc's Trigonometric Lemma
- Another Property of Points on Incircle
- Much from Little
- The Law of Cosines and the Law of Sines Are Equivalent
- Wonderful Trigonometry In Equilateral Triangle
- A Trigonometric Observation in Right Triangle
- A Quick Proof of cos(pi/7)cos(2.pi/7)cos(3.pi/7)=1/8

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