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Place the trapezoid in the coordinate plane with one base on the y-axis.
See solution.
We are asked to prove Theorem 6.24, the Trapezoid Midsegment Theorem. We are asked to write a coordinate proof, so let's place the trapezoid on the Coordinate Plane with the parallel bases parallel to the y-axis.
Let's place F at the origin. Since AF and CD are vertical, the x-coordinate of A is 0, and the x-coordinates of C and D are the same. Let's use the Midpoint Formula to express the coordinates of B and E in terms of the coordinates of the vertices.
The midpoint between points (x_1,y_1) and (x_2,y_2) is ( x_1+x_22, y_1+y_22). |
Let's substitute the coordinates of the vertices in this formula.
Points | Substitution | Midpoint |
---|---|---|
A(0,a) and C(p,c) | (0+p/2,a+c/2) | B(p/2,a+c/2) |
F(0,0) and D(p,d) | (0+p/2,0+d/2) | E(p/2,d/2) |
Since the first coordinates of B and E are the same, they lie on a vertical line. This proves that the midsegment BE is parallel to the bases.
To compare the length of the bases and the midsegment, we can use the fact that these segments are vertical, so the length is the difference between the y-coordinates of the endpoints.
Points | Distance |
---|---|
A(0,a) and F(0,0) | AF=a-0=a |
B(p/2,a+c/2) and E(p/2,d/2) | BE=a+c/2-d/2=a+c-d/2 |
C(p,c) and D(p,d) | CD=c-d |
Substitute expressions
LHS * 2=RHS* 2
Commutative Property of Addition