Let ABCD be a quadrilateral. Suppose AB and CD have equal length and angles BAD and BCD are supplementary (i.e., angle BAD plus angle BCD equals 180 degrees). Show that AD is parallel to BC.
Given: Quadrilateral ABCD, AB = CD, Angle BAD + Angle BCD = 180
Prove: Segment AD is parallel to Segment BC.
Assume not: Segment AD is NOT parallel to Segment BC.
Therefore, Line AB will intersect Line DC at some point, say F.
Let angle BAD = x and angle BCD= 180-x
Two Cases
I.) F is "outside" the line
Without loss of generality, B-A-F and C-D-F.
Note triangles BFC and AFD exist.
Angle FAD = 180-x, because angle BAD and angle FAD are a linear pair. Let angle BFD = a and angle ADF = b. Therefore, a + b = x.
Angle BCF = 180-x
Angle AFD = a
Therefore angle FBC = b.
Thus, triangle BCF is similar to trnagle DAF.
By definition of similarity, FD/(AB+FA) = FA/(CD+FD).
Let, AB=CD=y, FA=c, and FD=d. Note y,c,d >0. Thus c +d > 0
Thus, d/y+c = c/y+d
dy + d^2 = cy + c^2
y(d-c) = c^2 - d^2
y = - (c + d)
CONTRADICTION, y, c+d > 0, so they cannot be opposites.
II.) F is "inside" the line
Without loss of generality, A-B-F and D-F-C.
Note triangle AFD exists.
Therefore, x + angle BFD + angle ADF = 180.
Triangle BFC exists. Let angle CBF = a. (a > 0) Recall that angle BCF = 180 - x
Therfore, angle BFD = 180 - x + a, because the exterior angle is the sum of the two opposite interior angles in a triangle.
Therfore, x + 180 - x + a = 180
180 + a = 180
a = 0
CONTRADICATION. Because triangle BFC exists, a > 0.
The two cases cannot happen. Therefore, our assumption is wrong, and thus, segment AD is parallel to segment BC. Q.E.D.
Solution by Contradiction
