Midsegment Theorem and Coordinate Proof Study Guide

Biblical Integration: "For there is one God and one mediator between God and mankind, the man Christ Jesus." — 1 Timothy 2:5 (NIV)
Learning Objectives:
- Use properties of the Midsegment Theorem.
- Write coordinate proofs.

Theorem Statement: The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long as that side.
Proof References: Example 5 (p. 297); Exercise 41 (p. 300).
Mathematical Formulation:
- For a triangle $\Delta ABC$ with points $D$ and $E$ as the midpoints of sides $AB$ and $BC$ :
- $DE \parallel AC$
- $DE = \frac{1}{2} AC$ (or $AC = 2 \times DE$ )

Example 1: Roof Truss Construction
- Context: Triangles are utilized for structural strength in roof trusses.
- Given: $UV$ and $VW$ are midsegments of $\Delta RST$ .
- Given Measurements: $RT = 90\,\text{in.}$ and $VW = 57\,\text{in.}$
- Task: Find the lengths of $UV$ and $RS$
- Calculation of $UV$ :
  - Since $UV$ is the midsegment parallel to $RT$ :
  - $UV = \frac{1}{2} RT$
  - $UV = \frac{1}{2}(90\,\text{in.}) = 45\,\text{in.}$
- Calculation of $RS$ :
  - Since $VW$ is the midsegment parallel to $RS$ :
  - $RS = 2 \times VW$
  - $RS = 2(57\,\text{in.}) = 114\,\text{in.}$
Example 2: Geometric Proof in a Kaleidoscope Image
- Given: $AE = BE$ and $AD = CD$ in $\Delta ABC$
- Task: Show that $CB \parallel DE$
- Proof Steps:
  - Since $AE = BE$ , point $E$ is the midpoint of segment $AB$
  - Since $AD = CD$ , point $D$ is the midpoint of segment $AC$
  - Segment $DE$ connects the midpoints of two sides of $\Delta ABC$ , making $DE$ a midsegment of $\Delta ABC$
  - By Theorem 5.1 (Midsegment Theorem), a midsegment connecting two sides of a triangle is parallel to the third side
  - Therefore, $CB \parallel DE$

Definition: A coordinate proof involves placing geometric figures in a coordinate plane to prove geometric properties.
Generality: When variables are used to represent the coordinates of a figure in a coordinate proof, the resulting proof and formulas are true for all figures of that specific type.
Placement Strategies:
- Place at least one vertex at the origin $(0, 0)$ .
- Align at least one side along a coordinate axis (the x-axis or y-axis).
- Position the figure in the first quadrant whenever possible to maintain positive coordinate values.
- Assign variable coordinates that simplify calculations (such as using even multiples like $2a$ or $2b$ when midpoints are evaluated).

Example 3: Placing Figures for Side Length Calculations
- Part A: Placing a Rectangle
  - Position one vertex at the origin $(0, 0)$ .
  - Align the horizontal side along the positive x-axis and the vertical side along the positive y-axis.
  - Assigned Coordinates: $(0, 0)$ , $(h, 0)$ , $(h, k)$ , and $(0, k)$
- Part B: Placing a Scalene Triangle
  - Position one vertex at the origin $(0, 0)$ .
  - Align one side along the positive x-axis.
  - Assigned Coordinates: $(0, 0)$ , $(a, 0)$ , and $(b, c)$
Example 4: Analyzing an Isosceles Right Triangle
- Placement and Setup:
  - Place the vertex of the right angle at the origin $(0, 0)$ .
  - Align the two congruent legs along the positive x-axis and positive y-axis.
  - Let $k$ represent the length of each leg (where $k > 0$ ).
  - Assigned Vertex Coordinates: $(0, 0)$ , $(k, 0)$ , and $(0, k)$
- Finding the Length of the Hypotenuse:
  - The hypotenuse connects endpoints $(k, 0)$ and $(0, k)$
  - Applying the Distance Formula:
  - $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
  - $d = \sqrt{(k - 0)^2 + (0 - k)^2}$
  - $d = \sqrt{k^2 + (-k)^2}$
  - $d = \sqrt{k^2 + k^2}$
  - $d = \sqrt{2k^2}$
  - $d = k\sqrt{2}$
- Finding Midpoint $M$ of the Hypotenuse:
  - Applying the Midpoint Formula to endpoints $(k, 0)$ and $(0, k)$ :
  - $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
  - $M = \left(\frac{k + 0}{2}, \frac{0 + k}{2}\right)$
  - $M = \left(\frac{k}{2}, \frac{k}{2}\right)$