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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 ΔABC\Delta ABC be denoted by A(a)A(\mathbf{a}), B(b)B(\mathbf{b}), and C(c)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 HH.

Step 1: Define the altitudes and their relationships

We are given that ADAD and BEBE are the altitudes of the triangle, where:

  • DD is the foot of the altitude from vertex AA onto side BCBC,
  • EE is the foot of the altitude from vertex BB onto side CACA.

From the properties of altitudes, we know:

  • ADBC\text{AD} \perp \text{BC},
  • BECA\text{BE} \perp \text{CA}.

We aim to show that the altitudes ADAD, BEBE, and CFCF are concurrent, meaning they all intersect at point HH.

Step 2: Perpendicularity and vector equations

Since ADBCAD \perp BC, the vector from AA to HH (the point of intersection of the altitudes) must be perpendicular to vector BCBC. Therefore, we have the following equation:

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

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

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

Similarly, for altitude BEBE, we know that BECABE \perp CA. Hence, we have:

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

This leads to the equation:

B(AC)=0BC=CB(Equation 2)\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:

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

This can be rewritten as:

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

This equation indicates that the point HH, 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 HH.

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


Key Formulas or Methods Used

  • Dot product of vectors: For two vectors u\mathbf{u} and v\mathbf{v}, the dot product is given by:
uv=u1v1+u2v2+u3v3\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3

Two vectors are perpendicular if their dot product is zero:

uv=0\mathbf{u} \cdot \mathbf{v} = 0
  • Perpendicularity condition: If two vectors are perpendicular, their dot product must be zero:
uvuv=0\mathbf{u} \perp \mathbf{v} \quad \Rightarrow \quad \mathbf{u} \cdot \mathbf{v} = 0

Summary of Steps

  1. Define the altitudes of the triangle ADAD and BEBE, and their corresponding perpendicularity conditions.
  2. Use the vector dot product to express the condition of perpendicularity for the altitudes.
  3. Combine the equations for the perpendicularity of the altitudes to show that the altitudes meet at a common point.
  4. Conclude that the altitudes are concurrent, meaning they meet at the orthocenter HH.