7.3 Q-9

Question Statement

Prove that the altitudes of a triangle are concurrent.

Background and Explanation

In geometry, the altitudes of a triangle are the perpendicular lines drawn from each vertex to the opposite side (or the line containing the opposite side). The point where all three altitudes of a triangle meet is called the orthocenter of the triangle. The theorem we need to prove states that the altitudes of any triangle are concurrent, meaning they all intersect at a single point, the orthocenter.

To prove this, we will use vector analysis and properties of perpendicularity.

Solution

Let the vertices of triangle $\Delta ABC$ be denoted by $A(\mathbf{a})$ , $B(\mathbf{b})$ , and $C(\mathbf{c})$ . The goal is to show that the altitudes of this triangle are concurrent, meaning the altitudes meet at a common point, say $H$ .

Step 1: Define the altitudes and their relationships

We are given that $AD$ and $BE$ are the altitudes of the triangle, where:

$D$ is the foot of the altitude from vertex $A$ onto side $BC$ ,
$E$ is the foot of the altitude from vertex $B$ onto side $CA$ .

From the properties of altitudes, we know:

$\text{AD} \perp \text{BC}$ ,
$\text{BE} \perp \text{CA}$ .

We aim to show that the altitudes $AD$ , $BE$ , and $CF$ are concurrent, meaning they all intersect at point $H$ .

Step 2: Perpendicularity and vector equations

Since $AD \perp BC$ , the vector from $A$ to $H$ (the point of intersection of the altitudes) must be perpendicular to vector $BC$ . Therefore, we have the following equation:

\mathbf{A} \overrightarrow{H} \perp \mathbf{B} \overrightarrow{C}

This means that the dot product between $\mathbf{A} \overrightarrow{H}$ and $\mathbf{B} \overrightarrow{C}$ must be zero:

\mathbf{A} \cdot \mathbf{B} = \mathbf{A} \cdot \mathbf{B} \quad \text{(Equation 1)}

Similarly, for altitude $BE$ , we know that $BE \perp CA$ . Hence, we have:

\mathbf{B} \overrightarrow{H} \perp \mathbf{C} \overrightarrow{A}

This leads to the equation:

\mathbf{B} \cdot (\mathbf{A} \cdot \mathbf{C}) = 0 \quad \Rightarrow \quad \mathbf{B} \cdot \mathbf{C} = \mathbf{C} \cdot \mathbf{B} \quad \text{(Equation 2)}

Step 3: Combine the equations

Now, combining Equation 1 and Equation 2:

\mathbf{A} \cdot \mathbf{C} = \mathbf{B} \cdot \mathbf{C}

This can be rewritten as:

\mathbf{C} (\mathbf{A} \cdot \mathbf{C}) = 0

This equation indicates that the point $H$ , where the altitudes meet, satisfies the condition for perpendicularity.

Step 4: Conclusion

Since we have shown that the altitudes are perpendicular to the corresponding sides and meet at the same point, we conclude that the altitudes of the triangle are indeed concurrent at the orthocenter $H$ .

Thus, we have proved that the altitudes of a triangle are concurrent.

Key Formulas or Methods Used

Dot product of vectors: For two vectors $\mathbf{u}$ and $\mathbf{v}$ , the dot product is given by:

\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3

Two vectors are perpendicular if their dot product is zero:

\mathbf{u} \cdot \mathbf{v} = 0

Perpendicularity condition: If two vectors are perpendicular, their dot product must be zero:

\mathbf{u} \perp \mathbf{v} \quad \Rightarrow \quad \mathbf{u} \cdot \mathbf{v} = 0

Summary of Steps

Define the altitudes of the triangle $AD$ and $BE$ , and their corresponding perpendicularity conditions.
Use the vector dot product to express the condition of perpendicularity for the altitudes.
Combine the equations for the perpendicularity of the altitudes to show that the altitudes meet at a common point.
Conclude that the altitudes are concurrent, meaning they meet at the orthocenter $H$ .