Parallel lines exist in the same plane but do not intersect. Parallel lines have the same slope.
In diagrams, we usually indicate that two or more lines are parallel by placing an arrow symbol on each line, as shown. There are two sets of parallel lines (one consisting of three parallel lines, the other of two), and we use different arrow symbols to differentiate them, much as we'd use hash marks for congruent line segments.
We also use this written notation to match the diagram:
a || b, a || c, b || c, and d || e.
A transversal is a line or segment that cuts across two or more parallel lines, forming eight or more angles that we'll name below.
In this figure, the transversal, t, cuts parallel lines a and b.
In this figure, the pairs of angles formed by a transversal and parallel lines are shown. They are:
Those pairs are listed in the table below.
|Alternate interior angles: 4‒6, 3‒5||Angles between || lines and on opposite sides of the transversal|
|Alternate exterior angles: 1‒7, 2‒8||Angles on the outside of || lines and on opposite sides of the transversal|
|Corresponding angles: 1‒5, 4‒8, 2‒6, 3‒7||Angles in corresponding positions of their respective intersections, e.g. upper-left to upper left (1‒5)|
|Interior angles on the same side: 4‒5, 3‒6||Angles between || lines and on the same side of the transversal|
|Exterior angles on the same side: 1‒8, 2‒7||Angles on the outside of || lines and on the same side of the transversal|
|Vertical angles: 1‒3, 2‒4, 5‒7, 6‒8||Angles opposite one-another at intersections, e.g. 1‒3|
← This one is a postulate, something that's self-evident, and not possible to actually prove:
When two parallel lines are cut by the same line, any pair of corresponding angles formed are congruent. It makes sense, doesn't it? If the lines cut by the transversal are parallel, then the intersections produced are "clones" of one another.
The figure on the right shows two intersecting lines with four labeled angles. ∠A is supplementary to ∠B, and ∠D supp. ∠B.
If ∠A + ∠B = 180˚, then ∠A = 180˚ - ∠B and
if ∠D + ∠B = 180˚, then ∠D = 180˚ - ∠B,
so ∠A ≅∠D: Vertical angles are congruent.
← ∠1 ≅∠5 by the corresponding angle postulate, and ∠1 ≅∠3 because vertical angles are congruent, therefore ∠3 ≅∠5 by substitution of ∠3 for ∠1 in the first expression. Therefore alternate interior angles are congruent.
← ∠1 ≅∠5 by the corresponding angle postulate, and ∠5 ≅∠7 because vertical angles are congruent, therefore ∠1 ≅∠7 by substitution of ∠7 for ∠5 in the first expression. Therefore alternate exterior angles are congruent.
Using the numbered diagram above, we see that ∠1 ≅∠5 by the corresponding angle postulate, and ∠4 is supplementary to ∠1 (obvious from the diagram). That means ∠4 supp. ∠5 because an angle supplementary to a second angle must be supplementary to any other angle of the same measure.
Using the same numbered diagram, we see that ∠1 ≅∠5 by the corresponding angle postulate, and ∠5 supp. ∠8 (obvious from the diagram), so ∠1 supp. ∠8 because an angle supplementary to a second angle must be supplementary to any other angle of the same measure.
Now if you take a look back at all six proofs, you'll notice something interesting:
Any pair of angles formed by the intersection of a transversal with two parallel lines is either congruent or supplementary.
The parallel postulate says that through any given point, there is one and only one line parallel to another specific line.
Take a look at this figure. We can draw only one line through point P that is parallel to line L. This is one of the five postulates in Euclid's original writing on geometry. It's a postulate or axiom, because it can't actually be proved.
One way to rationalize it to yourself is just to consider line M || L. Now move M closer and closer to point P until P is on M. We already said that M || L, so it still is, and if we rotate it around point P the slightest bit, it will no longer be parallel.
This figure shows how we can use parallel lines to prove that the angles of a triangle sum to 180˚ (or π radians).
First we draw the line through the top angle of the triangle that is parallel to the base. There is only one such line by the parallel postulate. Then we use the complimentarity of angles e, b and d, along with alternate interior angles to prove that the sum of angles a, b and c must be the same as the sum of angles e, b and d, which is 180˚ because those angles are obviously complimentary.
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