7.3 Q-8

Question Statement

Prove that the perpendicular bisectors of the sides of a triangle are concurrent.

---

Background and Explanation

In geometry, the perpendicular bisectors of the sides of a triangle are the lines that are perpendicular to each side and pass through the midpoint of each side. A well-known result in geometry is that the perpendicular bisectors of the sides of any triangle are concurrent, meaning they meet at a single point. This point of intersection is called the circumcenter, and it is equidistant from all three vertices of the triangle.

To prove this, we will use vectors and the properties of perpendicularity, focusing on showing that the intersection of the perpendicular bisectors satisfies certain geometric conditions.

---

Solution

Let triangle $\Delta ABC$ have vertices $A(\mathbf{a})$, $B(\mathbf{b})$, and $C(\mathbf{c})$. Denote the midpoints of the sides as $D$, $E$, and $F$, corresponding to the sides $AB$, $BC$, and $CA$, respectively. We need to show that the perpendicular bisectors of these sides intersect at a common point.

Step 1: Perpendicular bisector of side $BC$

The midpoint of side $BC$ is $D$. The vector from $B$ to $C$ is $\mathbf{c} - \mathbf{b}$, and the vector from $B$ to $D$ is $\frac{1}{2}(\mathbf{c} + \mathbf{b})$. The perpendicular bisector of $BC$ is perpendicular to this vector. Hence, the dot product of $\frac{1}{2}(\mathbf{c} + \mathbf{b})$ and $\mathbf{c} - \mathbf{b}$ must be zero:

$$\frac{1}{2} (\mathbf{c} + \mathbf{b}) \cdot (\mathbf{c} - \mathbf{b}) = 0$$

Expanding this:

$$\frac{1}{2} (c^2 - b^2) = 0 \quad \Rightarrow \quad c^2 - b^2 = 0$$

Thus, we have:

$$c^2 = b^2 \tag{1}$$

Step 2: Perpendicular bisector of side $CA$

Similarly, the midpoint of side $CA$ is $E$. The vector from $A$ to $C$ is $\mathbf{c} - \mathbf{a}$, and the vector from $A$ to $E$ is $\frac{1}{2} (\mathbf{a} + \mathbf{c})$. The perpendicular bisector of $CA$ is perpendicular to this vector. Hence, the dot product of $\frac{1}{2} (\mathbf{a} + \mathbf{c})$ and $\mathbf{a} - \mathbf{c}$ must also be zero:

$$\frac{1}{2} (\mathbf{a} + \mathbf{c}) \cdot (\mathbf{a} - \mathbf{c}) = 0$$

Expanding this:

$$\frac{1}{2} (a^2 - c^2) = 0 \quad \Rightarrow \quad a^2 - c^2 = 0$$

Thus, we have:

$$a^2 = c^2 \tag{2}$$

Step 3: Combine the equations

Now, adding the results from equations (1) and (2):

$$a^2 - b^2 = 0 \quad \Rightarrow \quad a^2 = b^2$$

Step 4: Conclusion

From equations (1), (2), and the above result, we see that the perpendicular bisectors of the sides of triangle $\Delta ABC$ are concurrent at a single point. This point is the circumcenter, and it is equidistant from all three vertices of the triangle.

Thus, we have proven that the perpendicular bisectors of the sides of a triangle are concurrent.

---

Key Formulas or Methods Used

• Perpendicular bisector condition: For a line to be the perpendicular bisector of a segment, the dot product of the vector from the midpoint to one end of the segment and the vector from the midpoint to the other end must be zero:

$$\frac{1}{2} (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) = 0$$

• Concurrence of lines: For three lines to be concurrent, they must intersect at a single point, which in this case is the circumcenter of the triangle.

• Distance between points: The points where the perpendicular bisectors intersect are equidistant from all vertices of the triangle.

---

Summary of Steps

1. Calculate the vectors representing the sides of the triangle.
2. Find the midpoints of the sides of the triangle.
3. Use the perpendicularity condition (dot product equals zero) to establish relationships between the vectors.
4. Solve the equations derived from the perpendicular bisector conditions.
5. Conclude that the perpendicular bisectors are concurrent, as they intersect at the circumcenter.