Math 8 · G.CO.C.9

Corresponding angles

One line cuts across two parallel lines. How much does the angle at the first crossing tell you about the angle at the second?

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Alternate and same side anglesMath 8 Distance from a point to a lineGeometry

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Alternate and same side angles

Corresponding angles

The parallel postulate Distance from a point to a line

Why this exists

Two lines that never meet are hard to work with directly. You cannot measure an angle between them, because there is no angle between them. Nothing connects one to the other.

A third line that crosses both, a transversal, is the connection. It makes a crossing on each line, four angles at each, eight in all. The question is how the two groups of four are related.

When the two lines are parallel, the answer is as strong as it could be: the two crossings are copies of each other. Every angle at the first crossing reappears, at the same size and in the same position, at the second. That single fact is how an angle gets carried from one place in a figure to another, and most of what is known about the angles of a triangle and of a parallelogram is built on it.

Definition

A transversal is a line that crosses two other lines at two different points. In Figure 1 the transversal t crosses line l and line m, making four angles at each crossing. They are numbered the usual way: upper left, upper right, lower left, lower right, so \angle 1 to \angle 4 at l and \angle 5 to \angle 8 at m.

Figure 1Eight angles at two crossings. \angle 2 and \angle 6 sit in the same position, so they are corresponding angles.

Two of the eight angles are corresponding angles when they sit in the same position at the two crossings: on the same side of the transversal, and on the same side of their own line. \angle 2 is above l and to the right of t; \angle 6 is above m and to the right of t. They correspond. The four corresponding pairs are

\angle 1 \text{ and } \angle 5, \qquad \angle 2 \text{ and } \angle 6, \qquad \angle 3 \text{ and } \angle 7, \qquad \angle 4 \text{ and } \angle 8

The small arrowheads on l and m are the mark for parallel lines, and l \parallel m is read “l is parallel to m”. Being corresponding is only about position, so two lines that are not parallel still have corresponding angles. What parallel lines add is that corresponding angles are equal.

Intuition

Stand at the crossing on l and look at the four angles there. Now walk down the transversal to the crossing on m. If m runs in exactly the same direction as l, nothing about the view has changed: the transversal still leaves at the same tilt, and the line still runs off left and right the same way. The second crossing is the first one, moved.

Figure 2Slide the crossing at P down the transversal until P lands on Q. Line l lands on m, and the shaded angle lands on its partner.

That is the whole idea. Sliding the top crossing down t, without turning it, carries each of its four angles onto the angle in the same position below. Sliding never changes the size of an angle, so each pair of corresponding angles is equal.

If m were tilted even slightly, the view at the second crossing would be turned by that tilt, and the corresponding angles would differ by exactly that amount. Parallel is precisely the condition under which there is no tilt.

Results

Corresponding Angles Theorem. If two parallel lines are cut by a transversal, then each pair of corresponding angles is congruent. In Figure 1, if l \parallel m,

\angle 1 \cong \angle 5 \qquad \angle 2 \cong \angle 6 \qquad \angle 3 \cong \angle 7 \qquad \angle 4 \cong \angle 8

Converse. If two lines are cut by a transversal and one pair of corresponding angles is congruent, then the lines are parallel.

Put together with vertical angles and linear pairs, the theorem pins down all eight angles from one. At each crossing there are only two sizes, and they add to 180^\circ; corresponding angles copy those two sizes to the other crossing. So when l \parallel m,

m\angle 1 = m\angle 4 = m\angle 5 = m\angle 8 \qquad\text{and}\qquad m\angle 2 = m\angle 3 = m\angle 6 = m\angle 7

m\angle 1 + m\angle 2 = 180^\circ

The other named pairs follow in one step each. Alternate interior angles such as \angle 3 and \angle 6 are congruent, because \angle 3 \cong \angle 7 and \angle 7 is vertical to \angle 6. Alternate exterior angles are congruent in the same way, and same side interior angles such as \angle 4 and \angle 6 add to 180^\circ.

Why it is true

The argument is the slide from the intuition, made careful. It uses three facts about a translation, a slide of the whole plane that moves every point the same distance in the same direction:

it does not change the size of any angle;
it carries every line to a line parallel to it, or to the line itself;
it keeps every ray pointing the same way.

Call the two crossing points P on l and Q on m, and slide the plane so that P moves to Q, as in Figure 2.

The transversal slides along itself, since P and Q both lie on it. The ray of t going up from P becomes the ray of t going up from Q.

Line l slides to some line through Q that is parallel to l. But m is a line through Q parallel to l, and through a point off a line there is only one parallel to it. That is the parallel postulate. So the image of l is m, and the ray of l going right from P becomes the ray of m going right from Q.

Both sides of \angle 2 have now landed on the two sides of \angle 6. The slide carried \angle 2 exactly onto \angle 6, and a slide keeps angle sizes, so m\angle 2 = m\angle 6. The same slide carries each angle at P onto the angle in the same position at Q, which proves all four pairs at once.

The converse runs the same slide with the roles swapped. Suppose \angle 2 \cong \angle 6, and slide P to Q again. The right-hand ray of l lands on a ray from Q that makes the same angle with t, on the same side, as \angle 2 does; since \angle 6 is that size, the ray is the right-hand ray of m. So the image of l is m, and the image of a line under a translation is parallel to it. Hence l \parallel m.

Many geometry courses take the Corresponding Angles Theorem as a postulate instead of proving it, and then prove the alternate angle facts from it. Either way the parallel postulate is underneath, as the step above shows: it is the only reason the image of l has to be m.

Worked examples

Example 1

Finding eight angles from one

In Figure 3, l \parallel m. Find the angles marked a to g.

Figure 3One measure is given. Find the other seven.

Solution

Start at the crossing where the angle is known. b is vertical to the 65^\circ angle, so b = 65^\circ. The angles a and c each form a linear pair with the 65^\circ angle, so each is 180^\circ - 65^\circ = 115^\circ.

Now cross to line m. Each angle there corresponds to one at l, and l \parallel m, so each is equal to its partner:

d = a = 115^\circ \qquad e = 65^\circ \qquad f = b = 65^\circ \qquad g = c = 115^\circ

Check. The four angles at m go once around a point: 115 + 65 + 65 + 115 = 360.

Example 2

Solving for a variable in a corresponding pair

In Figure 4, l \parallel m. Find x, and the measure of each marked angle.

Figure 4A corresponding pair written as expressions in x.

Solution

The two marked angles are both above their line and to the right of t, so they are corresponding angles. The lines are parallel, so the angles are equal:

\begin{aligned} 3x + 15 &= 5x - 25 \\ 40 &= 2x \\ x &= 20 \end{aligned}

Substitute into both expressions, since they must agree:

3(20) + 15 = 75 \qquad\qquad 5(20) - 25 = 75

Each marked angle measures 75^\circ.

Example 3

Corresponding, then a linear pair

Use the numbering of Figure 1, with l \parallel m. If m\angle 2 = (2y + 20)^\circ and m\angle 8 = (4y + 10)^\circ, find y and both angles.

Solution

\angle 2 and \angle 8 are not a named pair, so find a path between them. \angle 2 corresponds to \angle 6, so m\angle 6 = (2y + 20)^\circ. At line m, \angle 6 and \angle 8 sit side by side on the right of t and form a linear pair, so they add to 180^\circ:

\begin{aligned} (2y + 20) + (4y + 10) &= 180 \\ 6y + 30 &= 180 \\ 6y &= 150 \\ y &= 25 \end{aligned}

Then m\angle 2 = 2(25) + 20 = 70^\circ and m\angle 8 = 4(25) + 10 = 110^\circ.

Setting the two expressions equal would have been wrong. They are not corresponding, and 70^\circ and 110^\circ are not equal. Deciding the relationship comes before any algebra.

Example 4

Deciding whether two lines are parallel

A transversal crosses lines p and q, with the angles numbered as in Figure 1. At p, m\angle 2 = 58^\circ. At q, m\angle 8 = 122^\circ. Are p and q parallel?

Solution

The converse says it is enough to find one pair of corresponding angles that are equal. The partner of \angle 8 is \angle 4, which is not given, but \angle 4 forms a linear pair with \angle 2:

m\angle 4 = 180^\circ - 58^\circ = 122^\circ

So m\angle 4 = m\angle 8 = 122^\circ. A pair of corresponding angles is congruent, so by the converse p \parallel q.

Had \angle 8 measured 120^\circ instead, the corresponding angles would differ, and the lines could not be parallel: if they were, the theorem would force the two angles to be equal.

Example 5

Finding the value that makes lines parallel

With the angles numbered as in Figure 1, m\angle 1 = (5x + 10)^\circ and m\angle 7 = (3x + 10)^\circ. For what value of x are the two lines parallel?

Solution

The lines are parallel exactly when a pair of corresponding angles is equal. \angle 7 corresponds to \angle 3, and \angle 3 forms a linear pair with \angle 1, so

m\angle 3 = 180 - (5x + 10) = 170 - 5x

Set the corresponding pair equal:

\begin{aligned} 170 - 5x &= 3x + 10 \\ 160 &= 8x \\ x &= 20 \end{aligned}

When x = 20, m\angle 1 = 110^\circ, m\angle 3 = 70^\circ and m\angle 7 = 70^\circ, so \angle 3 \cong \angle 7 and the lines are parallel. For any other x they are not.

Example 6

The angles of a parallelogram

ABCD is a parallelogram, so \overline{AD} \parallel \overline{BC} and \overline{AB} \parallel \overline{DC}, and m\angle A = 70^\circ. Using Figure 5, find the other three angles.

Figure 5Two sides extended past B turn each side of the parallelogram into a transversal.

Solution

Side \overline{AB} has been extended to E, so line AE is a transversal of the parallel lines AD and BC. \angle DAB and \angle CBE are both above line AE and to the right of their own line, so they correspond:

m\angle CBE = m\angle DAB = 70^\circ

\angle ABC and \angle CBE form a linear pair along line AE, so m\angle ABC = 180^\circ - 70^\circ = 110^\circ.

Now use the other pair of sides. Line CF is a transversal of the parallel lines AB and DC. \angle BCD, at C, lies below line DC and to the left of CF; the angle in the same position at B is \angle ABF. \angle ABF is vertical to \angle CBE, so it is 70^\circ, and therefore

m\angle BCD = m\angle ABF = 70^\circ

The same reasoning at the fourth corner, or the fact that the four angles of a quadrilateral add to 360^\circ, gives m\angle D = 110^\circ.

So opposite angles of a parallelogram are equal, and neighboring angles add to 180^\circ. The parallel sides did all the work, through corresponding angles.

Boundaries

The lines must be parallel for the angles to be equal. Corresponding angles exist for any two lines and any transversal, since the word only describes position. In Figure 6 the positions match but the sizes are 60^\circ and 45^\circ.

Figure 6Without parallel lines the positions still correspond, but the sizes do not.

The theorem and its converse are different statements. The theorem starts from parallel lines and concludes equal angles; the converse starts from equal angles and concludes parallel lines. Each needs its own argument, as every theorem and its converse does, and a proof has to cite the one whose hypothesis it actually has.

Both angles come from the same transversal. If a figure has several transversals, a corresponding pair is formed by one of them crossing the two lines. Two angles that look alike but sit on different transversals need not be equal.

It is a fact of flat geometry. The proof rests on the parallel postulate. On a sphere, where the “lines” are great circles and every two of them meet, the converse fails outright: two meridians both cross the equator at 90^\circ, so their corresponding angles are equal, and yet they meet at the poles.

Where this goes next

Corresponding angles are the first of the angle pairs at a transversal, and the others follow from them in a line each: alternate interior angles, alternate exterior angles and same side interior angles. With all of them in hand, a whole figure of parallel lines can be solved from one or two given angles.

The converse is the working test for parallel lines. It is how lines are proved parallel in a proof, and it is exactly what the compass construction of a parallel line through a point does: it copies an angle so that a pair of corresponding angles match. A transversal meeting one of two parallels at a right angle meets the other at a right angle too, the perpendicular transversal theorem.

And the theorem is the starting point for the most used fact about triangles: the three angles of a triangle add to 180^\circ.

Exercises

Naming the pairs

Use Figure 1 for Exercises 1 to 4.

Name the angle that corresponds to \angle 3.
Name the angle that corresponds to \angle 6.
Are \angle 4 and \angle 5 corresponding angles? Explain.
List all four pairs of corresponding angles.

Finding measures

In Exercises 5 to 8, l \parallel m and the angles are numbered as in Figure 1.

m\angle 1 = 112^\circ. Find m\angle 5 and m\angle 7.
m\angle 8 = 133^\circ. Find m\angle 4, m\angle 1 and m\angle 2.
A transversal crosses two parallel lines, and one of the eight angles is a right angle. Find all eight angles.
m\angle 7 = 58^\circ. Find all eight angles.

Using algebra

In Exercises 9 to 12, l \parallel m and the angles are numbered as in Figure 1.

m\angle 2 = (5x + 8)^\circ and m\angle 6 = (7x - 20)^\circ. Find x and the measure of each angle.
m\angle 1 = (3x + 34)^\circ and m\angle 5 = (4x + 9)^\circ. Find x and the measure of each angle.
m\angle 4 = (9y - 4)^\circ and m\angle 6 = (4y + 2)^\circ. Find y and both angles.
m\angle 3 = (x^2 + x)^\circ and m\angle 7 = (7x + 16)^\circ. Find every value of x that works, and the angle measure each one gives.

Parallel or not?

In Exercises 13 to 16 a transversal crosses lines p and q, with the angles numbered as in Figure 1 (p in the place of l, q in the place of m).

m\angle 2 = 64^\circ and m\angle 6 = 64^\circ. Are p and q parallel? Explain.
m\angle 1 = 117^\circ and m\angle 5 = 115^\circ. Are p and q parallel? Explain.
m\angle 3 = 71^\circ and m\angle 8 = 109^\circ. Are p and q parallel? Explain.
m\angle 2 = (6x - 8)^\circ and m\angle 6 = (4x + 22)^\circ. What value of x makes p \parallel q, and what are the angles then?

Using the idea elsewhere

In parallelogram PQRS, m\angle P = (2x)^\circ and m\angle Q = (3x + 30)^\circ. Find x and all four angles of the parallelogram.

Reasoning

A transversal crosses two parallel lines. Explain why the eight angles take at most two different measures.
With l \parallel m as in Figure 1, use corresponding angles and vertical angles to show that \angle 3 \cong \angle 6.
Someone says, “Corresponding angles are always congruent.” Explain why this is false, and correct the statement.
In the slide argument under Why it is true, which step uses the fact that l and m are parallel? What goes wrong in that step if they are not?
Lines l and m are cut by a transversal, numbered as in Figure 1, with m\angle 2 = 70^\circ and m\angle 6 = 75^\circ. The lines are not parallel, so they meet. Do they meet to the left of t or to the right? Explain.

Answers

\angle 7. Both are below their line and to the left of t.
\angle 2. Both are above their line and to the right of t.
No. \angle 4 is below l and right of t; \angle 5 is above m and left of t. They differ in both positions. The partner of \angle 4 is \angle 8. (\angle 4 and \angle 5 are alternate interior angles.)
\angle 1 and \angle 5, \angle 2 and \angle 6, \angle 3 and \angle 7, \angle 4 and \angle 8.
m\angle 5 = 112^\circ and m\angle 7 = 68^\circ. \angle 5 corresponds to \angle 1; \angle 7 forms a linear pair with \angle 5, so it is 180^\circ - 112^\circ.
m\angle 4 = 133^\circ, m\angle 1 = 133^\circ, m\angle 2 = 47^\circ. \angle 4 corresponds to \angle 8, \angle 1 is vertical to \angle 4, and \angle 2 forms a linear pair with \angle 4, so it is 180^\circ - 133^\circ.
All eight are 90^\circ. The right angle’s vertical partner and both its neighbors are 90^\circ, since 180^\circ - 90^\circ = 90^\circ, and corresponding angles copy all four to the other crossing.
m\angle 2 = m\angle 3 = m\angle 6 = m\angle 7 = 58^\circ and m\angle 1 = m\angle 4 = m\angle 5 = m\angle 8 = 122^\circ. \angle 3 corresponds to \angle 7, \angle 6 is vertical to \angle 7, and the rest are 180^\circ - 58^\circ.
x = 14, and each angle is 78^\circ. Corresponding angles are equal: 5x + 8 = 7x - 20, so 2x = 28. Check: 5(14) + 8 = 78 and 7(14) - 20 = 78.
x = 25, and each angle is 109^\circ. From 3x + 34 = 4x + 9, x = 25. Check: 3(25) + 34 = 109 and 4(25) + 9 = 109.
y = 14, with m\angle 4 = 122^\circ and m\angle 6 = 58^\circ. \angle 4 corresponds to \angle 8, which forms a linear pair with \angle 6, so (9y - 4) + (4y + 2) = 180 and 13y = 182.
x = 8, giving 72^\circ, or x = -2, giving 2^\circ. Setting the pair equal, x^2 + x = 7x + 16, so x^2 - 6x - 16 = 0 and (x - 8)(x + 2) = 0. Both give a positive angle: 8^2 + 8 = 72 = 7(8) + 16, and (-2)^2 + (-2) = 2 = 7(-2) + 16.
Yes. \angle 2 and \angle 6 are corresponding and equal, so by the converse p \parallel q.
No. \angle 1 and \angle 5 are corresponding but unequal. If p and q were parallel the theorem would make them equal.
Yes. \angle 3 and \angle 4 form a linear pair, so m\angle 4 = 180^\circ - 71^\circ = 109^\circ = m\angle 8. \angle 4 and \angle 8 correspond, so p \parallel q.
x = 15, and both angles are 82^\circ. The lines are parallel when 6x - 8 = 4x + 22, so 2x = 30. Check: 6(15) - 8 = 82 and 4(15) + 22 = 82.
x = 30; m\angle P = m\angle R = 60^\circ and m\angle Q = m\angle S = 120^\circ. Neighboring angles of a parallelogram add to 180^\circ (Example 6), so 2x + 3x + 30 = 180 and 5x = 150. Opposite angles are equal.
At each crossing, vertical angles are equal and neighbors add to 180^\circ, so the four angles there take at most two sizes. Corresponding angles copy those same sizes to the other crossing, so no new size appears.
\angle 3 \cong \angle 7, because they are corresponding angles and l \parallel m. At line m, \angle 7 (lower left) and \angle 6 (upper right) are vertical angles, so \angle 7 \cong \angle 6. So \angle 3 \cong \angle 6.
Corresponding describes position only, and any two lines cut by a transversal have corresponding angles; in Figure 6 they measure 60^\circ and 45^\circ. The correct statement is: if two parallel lines are cut by a transversal, then corresponding angles are congruent.
The step that says the image of l is m. The slide carries l to the line through Q parallel to l, and that is m only because m is parallel to l. If m is not parallel, the image of l is a different line through Q, so \angle 2 lands on an angle between t and that other line, not on \angle 6.
To the left. Between the lines and on the left of t are \angle 3 and \angle 5. m\angle 3 = 70^\circ, vertical to \angle 2, and m\angle 5 = 180^\circ - 75^\circ = 105^\circ, a linear pair with \angle 6. They add to 175^\circ, while on the right \angle 4 and \angle 6 add to 110^\circ + 75^\circ = 185^\circ. Two lines close up on the side where the two angles between them add to less than 180^\circ; that is Euclid’s own form of the parallel postulate. The triangle they form has a third angle of 180^\circ - 175^\circ = 5^\circ.