7.1 Q-13

Question Statement

Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long.

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Background and Explanation

In this problem, we are asked to prove a property of a triangle. Specifically, the midline theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

To prove this, we will use vectors. First, we will define the position vectors of the vertices of the triangle. Then, we will compute the vector representing the segment joining the midpoints and show that it is parallel to the third side and half as long.

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Solution

Step 1: Define the vertices of the triangle

Let the vertices of the triangle be $A$, $B$, and $C$, with their respective position vectors denoted as:

• $\mathbf{a}$ for $A$,
• $\mathbf{b}$ for $B$,
• $\mathbf{c}$ for $C$.

Step 2: Define the midpoints of sides $AB$ and $AC$

The midpoints $E$ and $F$ of sides $AB$ and $AC$, respectively, have the following position vectors:

• The position vector of $E$, the midpoint of $AB$, is given by:
  $\mathbf{E} = \frac{\mathbf{a} + \mathbf{b}}{2}$

• The position vector of $F$, the midpoint of $AC$, is given by:
  $\mathbf{F} = \frac{\mathbf{a} + \mathbf{c}}{2}$

Step 3: Find the vector $E \vec{F}$

To find the vector $E \vec{F}$, we subtract the position vector of $E$ from the position vector of $F$: $E \vec{F} = \mathbf{F} - \mathbf{E}$

Substitute the expressions for $\mathbf{F}$ and $\mathbf{E}$: $E \vec{F} = \frac{\mathbf{a} + \mathbf{c}}{2} - \frac{\mathbf{a} + \mathbf{b}}{2}$

Step 4: Simplify the expression

Simplifying the right-hand side: $E \vec{F} = \frac{(\mathbf{a} + \mathbf{c}) - (\mathbf{a} + \mathbf{b})}{2}$
$E \vec{F} = \frac{\mathbf{c} - \mathbf{b}}{2}$

Thus, we have: $E \vec{F} = \frac{1}{2} (\mathbf{c} - \mathbf{b})$

Step 5: Show that $E \vec{F}$ is parallel to $B \vec{C}$

Notice that $B \vec{C}$ is the vector from $B$ to $C$, which is given by: $B \vec{C} = \mathbf{c} - \mathbf{b}$

Therefore: $E \vec{F} = \frac{1}{2} B \vec{C}$

This shows that the vector $E \vec{F}$ is parallel to $B \vec{C}$ and half as long, as required.

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Key Formulas or Methods Used

• Midpoint Formula:
  The position vector of the midpoint of a line segment joining two points $A$ and $B$ is:
  $\text{Midpoint} = \frac{\mathbf{a} + \mathbf{b}}{2}$

• Vector Subtraction:
  The vector between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is:
  $\overrightarrow{A B} = (x_2 - x_1, y_2 - y_1)$

• Parallel Vectors:
  Two vectors are parallel if one is a scalar multiple of the other.

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Summary of Steps

1. Define the position vectors of the vertices of the triangle.
2. Calculate the position vectors of the midpoints of sides $AB$ and $AC$.
3. Find the vector joining the midpoints $E$ and $F$ by subtracting their position vectors.
4. Simplify the vector expression and show that it is parallel to the vector $B \vec{C}$.
5. Conclude that the line segment joining the midpoints is parallel to the third side and half its length.