Proving A Midsegment In A Triangle ✓ Solved

Proving a Midsegment in a Triangle

Verify the midpoints of A(1, -3) B(1, 7) and C(9, 5). Verify that EF is parallel to BC. Show the segments have the same slope. Verify that BC is twice the distance of EF. Is EF the midsegment of triangle ABC? EXPLAIN your answer.

Paper For Above Instructions

In this paper, I aim to explore the concept of midsegments in triangles through specific examples and mathematical proofs. A midsegment in a triangle is defined as a line segment that connects the midpoints of two sides of the triangle. We will verify the midpoints of given points, establish the parallel relationship between said midsegment and the base of the triangle, and conclude whether the line segment qualifies as a midsegment based on its characteristics.

Step 1: Verification of Midpoints

The points provided are A(1, -3), B(1, 7), and C(9, 5). First, we need to find the midpoints of segments AB and AC. The formula for finding the midpoint M of a segment connecting two points (x₁, y₁) and (x₂, y₂) is given by:

M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)

Calculating the midpoint of AB:

M_AB = ((1 + 1) / 2, (-3 + 7) / 2) = (1, 2)

Next, we calculate the midpoint of AC:

M_AC = ((1 + 9) / 2, (-3 + 5) / 2) = (5, 1)

Step 2: Verification of Parallelism

Next, we need to demonstrate that segment EF (the segment connecting midpoints M_AB and M_AC) is parallel to BC. Two lines are parallel if they have the same slope. To find the slope of line BC, we use the slope formula:

Slope = (y₂ - y₁) / (x₂ - x₁)

Calculating the slope of segment BC:

Let B = (1, 7) and C = (9, 5). Then:

Slope_BC = (5 - 7) / (9 - 1) = (-2) / (8) = -1/4

Now, we calculate the slope of segment EF:

Let E = (1, 2) and F = (5, 1). Then:

Slope_EF = (1 - 2) / (5 - 1) = (-1) / (4) = -1/4

Since both slopes are equal, we conclude that EF is parallel to BC, verifying that EF || BC.

Step 3: Verification of Lengths

To demonstrate that BC is twice the length of EF, we first compute the lengths of segments BC and EF. The length of a segment can be determined using the distance formula:

Distance = √((x₂ - x₁)² + (y₂ - y₁)²)

Calculating the length of BC:

Length_BC = √((9 - 1)² + (5 - 7)²) = √(8² + (-2)²) = √(64 + 4) = √68 = 2√17

Now for EF:

Length_EF = √((5 - 1)² + (1 - 2)²) = √(4² + (-1)²) = √(16 + 1) = √17

We observe that:

Length_BC = 2 × Length_EF

Thus, BC is indeed twice the length of segment EF.

Conclusion

Based on the calculations and explorations above, we can affirmatively conclude that segment EF, connecting the midpoints of sides AB and AC, is indeed the midsegment of triangle ABC. EF is parallel to base BC and its length is precisely half that of BC, satisfying the properties of a midsegment. Consequently, we have successfully verified that EF is the midsegment of triangle ABC.

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