Coordinate Proof Using Distance With Segments And Triangles - Fill ... - Free Printable

You're working on a coordinate proof problem using the Midsegment Theorem. Let’s walk through each step carefully.

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## 📌 Problem Recap:

We are given an isosceles triangle ABC with side lengths:
- AB = 8 units
- AC = BC = 5 units

We’re to place this triangle in the coordinate plane so that:
- The long side (AB) is centered on the x-axis at the origin.
- Then, we’ll find coordinates of midpoints D, E, F of sides AB, BC, and AC respectively.
- Use the distance formula to show DE = ½ AB.
- Finally, explain how to prove that segment FE is a midsegment and what other steps are needed.

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## ✔ Step 1: Place Triangle ABC on Coordinate Plane

Since AB is the base (longest side, 8 units), and it’s centered on the x-axis at the origin:

- Let point A be at (-4, 0) → because from -4 to +4 is 8 units total.
- Let point B be at (4, 0) → symmetric about origin.

Now, point C must be such that AC = BC = 5 units → since it’s isosceles with AC = BC.

Let’s find coordinates of point C.

Assume C is at (0, h) — directly above the origin (since triangle is symmetric).

Use distance formula from A to C:

> Distance AC = √[(0 - (-4))² + (h - 0)²] = √[16 + h²] = 5

So:

> √(16 + h²) = 5
> Square both sides:
> 16 + h² = 25
> h² = 9
> h = 3 or h = -3 → Since it's a triangle above the base, use h = 3

✔ So, point C is at (0, 3)

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## ✔ Step 2: Find Midpoints D, E, F

Recall: Midpoint formula = ((x₁+x₂)/2, (y₁+y₂)/2)

- D = midpoint of AB → A(-4,0), B(4,0)
→ D = ((-4+4)/2, (0+0)/2) = (0, 0)

- E = midpoint of BC → B(4,0), C(0,3)
→ E = ((4+0)/2, (0+3)/2) = (2, 1.5)

- F = midpoint of AC → A(-4,0), C(0,3)
→ F = ((-4+0)/2, (0+3)/2) = (-2, 1.5)

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## ✔ Step 3: Use Distance Formula to Show DE = ½ AB

First, compute DE:

D = (0, 0), E = (2, 1.5)

> DE = √[(2 - 0)² + (1.5 - 0)²] = √[4 + 2.25] = √6.25 = 2.5

Now, AB = 8, so half of AB is 4? Wait — that doesn’t match!

Hold on — something’s off.

Wait — let’s double-check.

Actually, DE connects midpoints of AB and BC — so by the Midsegment Theorem, DE should be parallel to AC and half its length — NOT half of AB.

But the question says: “Use the distance formula to show equal side lengths: DE = ___”

Looking back at the worksheet, it says:

> “Use the distance formula to show equal side lengths: DE = ___ DF = ___”

And then: “Explain... you want to use ΔABC to prove the Midsegment Theorem, given that FE is a midsegment...”

Ah! I think there’s a misalignment in the instructions. Let me re-read.

Actually, looking at the diagram and the labels:

In the diagram provided (even though I can’t see it, but based on standard labeling):

- D is midpoint of AB
- E is midpoint of BC
- F is midpoint of AC

Then, FE connects midpoints of AC and BC → so FE is the midsegment parallel to AB.

And DE connects midpoints of AB and BC → so DE is midsegment parallel to AC.

So the problem likely wants us to show:

> DE = ½ AC
> DF = ½ BC
> FE = ½ AB

Because those are the midsegments.

But the worksheet says: “show DE = ___ DF = ___” — probably expecting numerical values.

Let’s compute all three distances to be thorough.

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Compute DE:

D = (0,0), E = (2, 1.5)
DE = √[(2-0)² + (1.5-0)²] = √[4 + 2.25] = √6.25 = 2.5

AC = 5 → so DE = ½ AC → ✔ Correct!

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Compute DF:

D = (0,0), F = (-2, 1.5)
DF = √[(-2-0)² + (1.5-0)²] = √[4 + 2.25] = √6.25 = 2.5

BC = 5 → so DF = ½ BC → ✔ Correct!

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Compute FE:

F = (-2, 1.5), E = (2, 1.5)
FE = √[(2 - (-2))² + (1.5 - 1.5)²] = √[16 + 0] = √16 = 4

AB = 8 → so FE = ½ AB → ✔ Correct!

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## ✔ Step 4: Fill in the blanks

From above:

> DE = 2.5
> DF = 2.5
> FE = 4

Note: Since AC = BC = 5, and AB = 8, these make sense.

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## ✔ Step 5: Explain how to prove FE is a midsegment and what other steps are needed

> “Explain: Instead that you want to use ΔABC to prove the Midsegment Theorem, given that FE is a midsegment and FE = ½ AB. What other steps would you need to complete the coordinate proof?”

🔍 What is the Midsegment Theorem?

> The segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length.

We’ve already shown that FE = ½ AB (numerically).

But to fully prove the Midsegment Theorem using coordinate geometry, we also need to show that FE is parallel to AB.

➤ How to show parallelism?

Two lines are parallel if they have the same slope.

- Slope of AB: A(-4,0), B(4,0) → slope = (0 - 0)/(4 - (-4)) = 0/8 = 0

- Slope of FE: F(-2,1.5), E(2,1.5) → slope = (1.5 - 1.5)/(2 - (-2)) = 0/4 = 0

✔ So FE is parallel to AB.

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## ✔ Final Answer Summary:

Coordinates:

- A(-4, 0)
- B(4, 0)
- C(0, 3)
- D (mid AB): (0, 0)
- E (mid BC): (2, 1.5)
- F (mid AC): (-2, 1.5)

Distances:

- DE = 2.5
- DF = 2.5
- FE = 4

To Prove FE is a Midsegment:

We showed:
1. FE connects midpoints of AC and BC → by definition, it’s a midsegment.
2. FE = ½ AB (numerically verified).
3. FE is parallel to AB (slopes both 0).

✔ Therefore, we’ve proven the Midsegment Theorem for FE in triangle ABC using coordinate geometry.

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## 📝 Boxed Final Answers for Worksheet:

> D: (0, 0)
> E: (2, 1.5)
> F: (-2, 1.5)
> DE = 2.5
> DF = 2.5
> FE = 4

> Other steps needed: Show that FE is parallel to AB by computing slopes (both are 0), which confirms the full Midsegment Theorem (parallel + half-length).

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Let me know if you’d like to graph this or verify any step visually!