Geometry · G.CO.C.10

Indirect proof

Some facts in geometry say that something never happens. Two lines never cross twice, a triangle never has two right angles, no side of a triangle is as long as the other two together. How can a proof show that something never happens?

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Indirect proof

When you reach for it

Most proofs run forward, from the Given to the Prove. That stalls when the Prove says something does not happen. “\ell \parallel m” means the lines never meet, however far they run. “A triangle has at most one right angle” means a second one is never there. There is nothing to work on: no meeting point to name, no second right angle to measure.

An indirect proof supplies the missing object. It assumes, for the length of the argument, that the Prove is false: the lines do meet, at some point X. Then it reasons from the Given and the assumption, using only results already established, until it reaches something that cannot be true. Every step was valid and the Given is true, so the assumption failed, and the Prove holds.

Reach for it when the Prove says “not”, “never”, “at most one” or “parallel”, and its negation hands you a point, a line or an angle to name. The logic is already done: what the proof reaches is a contradiction, and proof by contradiction is the method in general. This lesson uses it in geometry.

The best-known fact of this shape is about lengths. Lay sticks of length 3, 4 and 8 on a table. The two short ones end to end reach only 7, so they cannot meet the far end of the long one, and no triangle closes. Every triangle behaves this way. The triangle inequality says that in a triangle ABC

AB < AC + CB

and the same for each of the other two sides: one side is always shorter than the other two together. The road from A to B by way of C is never shorter than the straight one.

A drawing cannot prove this. The claim is about every triangle, including the ones nobody has drawn, and it says that something never happens: no triangle has a side as long as the other two together. So it is a job for an indirect proof.

Here the assumption is unusually useful. It is a statement about lengths, and it says exactly where two points must sit on the long side. From those two points, angle facts do the rest. So the move is turning a fact about lengths into a fact about angles, because the tools on the table are the angle theorems. It is the move to make when the claim is that one length, or one sum of lengths, is always smaller than another, and the facts you already own are about angles. The fact itself is met first as a rule about sticks, stated but not proved, in the triangle inequality. The second half of the procedure below is why it is true, and Examples 7 to 13 put it to work.

The procedure

The four parts of an indirect proof

Start with the usual set-up from Given, Prove and the diagram. An indirect proof then has an assumption, an argument from it, a contradiction and a conclusion, and each part is there for a reason.

Assume the negation of the Prove. It must be the complete opposite, covering every way the Prove could fail (negation). The common ones:

The Prove says	The assumption says
\ell \parallel m	\ell and m meet at a point; call it X
\ell \nparallel m	\ell \parallel m
\angle A \cong \angle B	m\angle A \neq m\angle B
AB \neq AC	AB = AC
AB < AC + CB	AB \ge AC + CB
at most one angle is right	at least two angles are right
exactly one line does it	no line does it, or at least two do

The first row needs the lines in one plane, where two different lines either meet or are parallel. The last splits in two because “exactly one” is two claims. Write the assumption as its own sentence, “Assume, to reach a contradiction, that ...”, because every line after it depends on something the proof will throw away.

Reason forward from the Given and the assumption. Each step needs the usual reason: a given, the assumption, a definition, a postulate or a theorem already proved. Name and draw any object the assumption supplies. The drawing now shows something that cannot exist, so it is a sketch, and as in what a diagram may assume, nothing is read off it.

Stop at a contradiction, and name it. The target is two statements that cannot both be true. One has been derived; the other is a given, the assumption, or a known result. Say which.

Conclude the Prove. With G for the Given and P for the Prove, the proof has shown that G and \neg P cannot both hold. G does, so \neg P is false and P is true. Why that counts as proof is in the contradiction lesson: “if \neg P, then a contradiction” has the same truth table as P. It is modus tollens with the contradiction as the denied consequent.

Proving the triangle inequality indirectly

Here are the four parts at work on the triangle inequality. The proof uses four facts that have already been proved, all of them about angles: the base angles of an isosceles triangle are equal, the angle addition postulate, the linear pair theorem (two angles that fill a straight line add to 180^\circ), and the triangle angle sum (180^\circ).

Given. \triangle ABC, three points not on one line.

Prove. AB < AC + CB.

The triangle is any triangle, and any side can be called AB, so proving this for one side proves it for all three.

Assume the negation, all of it. The Prove is a strict “less than”, so its opposite is “at least as much as”, not “more than”. Assume, to reach a contradiction, that

AB \ge AC + CB

Writing “more than” would leave the case AB = AC + CB open. “At least” covers both, and the proof below handles both at once.

Turn the assumption into positions. Mark a point D on \overline{AB} with AD = AC, and a point E on \overline{AB} with BE = BC. Both exist, since the assumption makes AB longer than AC and longer than CB (the other length is positive), so each of the two lengths fits on the segment.

Measured from A, the point D is AC away and E is AB - BC away. The assumption, rearranged, says AB - BC \ge AC, so E is at least as far from A as D is. Along the line the order is A, D, E, B, and D and E may be the same point.

Figure 1The assumption, sketched. The dim ticks say AC = AD and BC = BE, and the order A, D, E, B is the assumption. No accurate drawing of it exists.

This is the only place the assumption is used. Figure 1 is a sketch, because the assumption cannot be drawn accurately, and nothing is read off it.

Get two pairs of equal angles. Since AC = AD, the triangle ACD is isosceles, and BC = BE makes BCE isosceles too. They are genuine triangles because C is not on the line AB, which is the Given. By the base angles theorem,

\begin{aligned} m\angle ACD &= m\angle ADC \\ m\angle BCE &= m\angle BEC \end{aligned}

Give the sum of the two base angles at D and E a short name:

s = m\angle ADC + m\angle BEC

By the equalities above, s is also the sum of the two base angles at C, m\angle ACD + m\angle BCE.

Count the angle at C. D and E are on the segment \overline{AB}, so they lie inside \angle ACB, and the rays from C come in the order CA, CD, CE, CB. The angle addition postulate, used twice, gives

m\angle ACB = m\angle ACD + m\angle DCE + m\angle ECB

where m\angle DCE = 0^\circ if D = E. The first and last terms add to s, so m\angle ACB = s + m\angle DCE, that is,

s = m\angle ACB - m\angle DCE \qquad (1)

Count the angles along the line. At D, the angles \angle ADC and \angle CDE form a linear pair, because A, D, E are on one line with D between A and E. At E, the angles \angle BEC and \angle CED form another. So the four angles add to 360^\circ, and

s = 360^\circ - (m\angle CDE + m\angle CED)

The triangle angle sum in \triangle CDE says m\angle CDE + m\angle CED = 180^\circ - m\angle DCE. Therefore

s = 180^\circ + m\angle DCE \qquad (2)

If D = E there is no triangle CDE, but then \angle ADC and \angle BDC are one linear pair, so s = 180^\circ, and (2) still holds with m\angle DCE = 0^\circ.

Put the two counts together. Both (1) and (2) give s, so their right sides are equal:

m\angle ACB - m\angle DCE = 180^\circ + m\angle DCE

Move m\angle DCE across:

m\angle ACB = 180^\circ + 2\,m\angle DCE \ge 180^\circ

Name the contradiction. \angle ACB is an angle of a triangle. The other two angles are positive and all three add to 180^\circ, so m\angle ACB < 180^\circ. The proof has derived m\angle ACB \ge 180^\circ, and the triangle angle sum says the opposite. They cannot both be true.

Conclude. Every step followed from the Given, the assumption and facts already proved, so the assumption is what failed. AB < AC + CB. Since AB was any side of any triangle, each side of a triangle is shorter than the other two together.

The contradiction has a picture. If AB were at least AC + CB, the two short sides, each swung down onto \overline{AB} from its own end, would at best just meet and could never overlap, and the angle at C would have to lie flat or bend past flat. A triangle whose angle at C is exactly flat is not a triangle at all, it is the segment \overline{AB} with C somewhere on it. That is the equality case, and Example 7 shows what happens to the same construction on a triangle that really exists.

Worked examples

Example 1

Two lines cross only once

Prove that two different lines meet in at most one point.

Solution

Given. \ell and m are lines, and \ell \neq m.

Prove. \ell and m have at most one point in common.

Plan. “At most one” fails only if there are two. Two points pin down a line: through two points there is exactly one line.

Figure 2The assumption, drawn. A second line through A and B can only be drawn by bending it.

Assume, to reach a contradiction, that \ell and m have two different points in common, A and B. Then \ell and m are both lines through A and B. Through two points there is exactly one line, so \ell = m, which contradicts the Given. So \ell and m share at most one point.

Figure 2 shows why a picture of the assumption can only be a sketch. The proof never looked at it. The fact belongs to intersections of lines and planes, and the later examples use it.

Example 2

Unequal corresponding angles

Lines \ell and m are cut by a transversal t, with the angles numbered as in corresponding angles: \angle 1 to \angle 4 at \ell and \angle 5 to \angle 8 at m, each group upper left, upper right, lower left, lower right. m\angle 2 = 71^\circ and m\angle 7 = 107^\circ. Prove that \ell \nparallel m.

Solution

Given. t crosses \ell and m; m\angle 2 = 71^\circ; m\angle 7 = 107^\circ.

Prove. \ell \nparallel m.

Plan. The negation is “parallel”, and parallel lines come with a theorem that fixes their angles. Find the partner of \angle 2 and compare.

Assume, to reach a contradiction, that \ell \parallel m. \angle 6 and \angle 7 are vertical angles, so m\angle 6 = 107^\circ. \angle 2 and \angle 6 are corresponding angles at parallel lines, so the corresponding angles theorem gives m\angle 2 = 107^\circ. That contradicts the Given, m\angle 2 = 71^\circ. So \ell \nparallel m.

Example 3

At most one right or obtuse angle

Prove that a triangle has at most one angle of 90^\circ or more.

Solution

Given. \triangle ABC.

Prove. At most one of \angle A, \angle B and \angle C measures 90^\circ or more.

Plan. Two big angles use up the whole 180^\circ.

Assume, to reach a contradiction, that two angles measure 90^\circ or more, and name the triangle so that they are \angle A and \angle B. Then m\angle A + m\angle B \ge 180^\circ, and by the triangle angle sum

m\angle C = 180^\circ - m\angle A - m\angle B \le 0^\circ

But every angle measures more than 0^\circ. So at most one angle measures 90^\circ or more. Angles of 95^\circ and 90^\circ, for instance, would leave 180^\circ - 95^\circ - 90^\circ = -5^\circ for the third. A triangle therefore has at most one right angle, at most one obtuse angle, and never one of each.

Example 4

Two perpendiculars to the same line

Prove that two different lines in a plane perpendicular to the same line are parallel.

Solution

Given. \ell, m and t lie in one plane; \ell \perp t at P; m \perp t at Q; \ell \neq m.

Prove. \ell \parallel m.

Plan. If \ell and m meet, the meeting point and the two feet make a triangle with two right angles, which Example 3 rules out. Note first that P \neq Q, since in a plane only one line is perpendicular to t at a given point of t.

Figure 3A sketch of the assumption. The Given makes \angle 1 and \angle 2 right angles, so no accurate drawing of this exists.

This time the proof is set out in two columns.

Statements	Reasons
1. \ell \perp t at P	given
2. m \perp t at Q	given
3. \ell \neq m	given
4. P \neq Q	in a plane only one line is perpendicular to t at a point of t: lines 1, 2 and 3
5. \ell and m meet at a point X	assumption, to reach a contradiction
6. X is not on t	two lines share at most one point (Example 1): on t, X would be both P and Q, lines 1, 2, 4 and 5
7. \triangle PQX is a triangle	definition of triangle, three points not on one line: line 6
8. \angle XPQ is a right angle	perpendicular lines make four right angles: line 1
9. \angle XQP is a right angle	perpendicular lines make four right angles: line 2
10. \triangle PQX has at most one right angle	a triangle has at most one right angle (Example 3): line 7
11. \ell and m do not meet	contradiction: lines 8, 9 and 10 cannot all hold, so line 5 is false
12. \ell \parallel m	definition of parallel lines: lines 3 and 11, in one plane

Figure 3 sketches statement 5. Its angles at P and Q look nothing like right angles, and cannot be made to: that is what the proof shows.

Example 5

Unequal angles, no straight line

Rays \overrightarrow{EA}, \overrightarrow{EB}, \overrightarrow{EC} and \overrightarrow{ED} start at E, and E is between C and D. m\angle AEC = 70^\circ and m\angle BED = 45^\circ. Prove that E is not between A and B.

Solution

Given. E is between C and D; m\angle AEC = 70^\circ; m\angle BED = 45^\circ.

Prove. E is not between A and B.

Plan. If A, E and B were a straight line through E, the two angles would be vertical angles, and the Vertical Angles Theorem would make them equal.

Assume, to reach a contradiction, that E is between A and B. Then \overrightarrow{EA} and \overrightarrow{EB} are opposite rays, and so are \overrightarrow{EC} and \overrightarrow{ED}, by the Given. So \angle AEC and \angle BED are vertical angles, and the Vertical Angles Theorem gives m\angle AEC = m\angle BED, that is, 70^\circ = 45^\circ. That contradicts the Given. So E is not between A and B: the rays \overrightarrow{EA} and \overrightarrow{EB} bend at E.

Look at what was used. The Given, 70^\circ \neq 45^\circ, says the theorem’s conclusion \angle AEC \cong \angle BED is false. A true conditional with a false conclusion has a false hypothesis, so “the angles are vertical” is false: that is modus tollens. Put another way, the proof applied the contrapositive of the Vertical Angles Theorem, “if two angles are not congruent, they are not vertical angles”, which is true because the theorem is. Written that way, with no assumption made and thrown away, it is a direct proof (Exercise 34).

Example 6

The converse of the corresponding angles theorem

Prove: if two lines are cut by a transversal and a pair of corresponding angles are congruent, then the lines are parallel.

Solution

Given. \ell, m and t lie in one plane; t crosses \ell at P and m at a different point Q, with P above Q; \angle 1 at P and \angle 2 at Q are corresponding angles, each above its line and right of t; \angle 1 \cong \angle 2.

Prove. \ell \parallel m.

Plan. If the lines meet, they make a triangle with t, and one of the two given angles becomes an exterior angle of it. Euclid’s Proposition 16, the exterior angle inequality, says an exterior angle of a triangle is larger than each of the two interior angles not next to it. Euclid proves it without the parallel postulate, as the parallel postulate lesson explains. Neither the postulate nor the corresponding angles theorem is used. Euclid proves this converse before he first calls on the postulate, and it is the theorem direction that needs it.

Figure 4Case 1 of the assumption, sketched. \angle 1 is an exterior angle of \triangle PQX and \angle 2 is one of its far interior angles, so the Given \angle 1 \cong \angle 2 cannot hold here.

Assume, to reach a contradiction, that \ell \nparallel m. The lines are different, since P is on \ell and not on m (t meets m only at Q, by Example 1). Different lines in one plane that are not parallel meet, say at X. X is not on t, or it would be P (where \ell meets t) and also Q. So P, Q and X form a triangle, and X lies on one side of t. That gives two cases (proof by cases).

X to the right of t (Figure 4). The ray \overrightarrow{PX} is the right-hand ray of \ell, and \overrightarrow{QX} the right-hand ray of m. So the triangle’s angle at Q, between \overrightarrow{QP} (up t) and \overrightarrow{QX}, is \angle 2. At P, the side \overline{QP} carries on beyond P as the ray of t going up, and the angle between that ray and \overrightarrow{PX} is \angle 1: an exterior angle of \triangle PQX. By the exterior angle inequality, m\angle 1 > m\angle 2.

X to the left of t. Now \overrightarrow{QX} is the left-hand ray of m, and side \overline{XQ} carries on beyond Q as the right-hand ray of m. So \angle 2 is the exterior angle, at Q. The triangle’s angle at P lies between the ray of t going down and the left-hand ray of \ell; it is vertical to \angle 1, so it has the same measure. The exterior angle inequality gives m\angle 2 > m\angle 1.

Either way the two angles are unequal, which contradicts \angle 1 \cong \angle 2. So \ell \parallel m. This is Euclid’s own proof, Book I, Proposition 27, told for corresponding angles. A theorem and its converse reaches the same converse through Proposition 17, and the corresponding angles lesson through a slide.

Example 7

The triangle inequality construction on a real triangle

\triangle ABC has m\angle A = m\angle B = 50^\circ, so m\angle ACB = 80^\circ. Carry out the construction of the triangle inequality proof: D on ray \overrightarrow{AB} with AD = AC, and E on ray \overrightarrow{BA} with BE = BC. Find the angles at C, say where D and E fall, and compare with what the assumption would force.

Solution

Figure 5The same construction on a real triangle. Now E falls between A and D, and \angle ECD = 50^\circ is counted twice.

\triangle ACD is isosceles with AC = AD. Its angle at A is \angle CAB = 50^\circ, since D is on ray \overrightarrow{AB}. The base angles are equal, and the three angles add to 180^\circ:

m\angle ACD = m\angle ADC = \frac{180^\circ - 50^\circ}{2} = 65^\circ

In the same way m\angle BCE = m\angle BEC = 65^\circ.

Now 65^\circ < 80^\circ, so ray \overrightarrow{CD} is inside \angle ACB, which puts D on the segment \overline{AB}. The same holds for E. Angle addition then gives

\begin{aligned} m\angle ACE &= m\angle ACB - m\angle BCE \\ &= 80^\circ - 65^\circ = 15^\circ \end{aligned}

which is less than m\angle ACD = 65^\circ. So ray \overrightarrow{CE} comes before \overrightarrow{CD}, and E lies between A and D. The order along the line is A, E, D, B: the reverse of the assumed order, and m\angle ECD = 65^\circ - 15^\circ = 50^\circ.

Now redo both counts with the order reversed. At C, the two base angles overlap along \angle ECD:

\begin{aligned} m\angle ACD + m\angle BCE &= 130^\circ \\ &= 80^\circ + 50^\circ \\ &= m\angle ACB + m\angle ECD \end{aligned}

Along the line, \angle ADC and \angle BEC are the angles at D and E of \triangle CED, so they add to 180^\circ - m\angle ECD = 130^\circ. Equating the two counts, m\angle ACB + m\angle ECD = 180^\circ - m\angle ECD, so

\begin{aligned} m\angle ACB &= 180^\circ - 2\,m\angle ECD \\ &= 180^\circ - 100^\circ = 80^\circ \end{aligned}

which is right. Compare the proof. With the assumed order it found 180^\circ + 2\,m\angle DCE, and on a real triangle it finds 180^\circ - 2\,m\angle ECD. The angle at C sits below 180^\circ when the points overlap and would sit at or above it if they did not. As a triangle is flattened, the overlap shrinks to nothing and the angle at C climbs to 180^\circ.

Example 8

Can these lengths be the sides of a triangle?

Decide for each set: (a) 6, 9, 14 (b) 5, 7, 13 (c) 4, 6, 10 (d) 2.5, 2.5, 4.9

Solution

Only the longest side can fail. Call the sides a \le b \le c. The other two tests are automatic, because the lengths are positive: a \le c < c + b and b \le c < c + a. So it is enough to compare the longest side with the sum of the other two.

(a) 6 + 9 = 15, and 14 < 15. Passes.

(b) 5 + 7 = 12, and 13 is not less than 12. Fails.

(c) 4 + 6 = 10, and 10 is not less than 10. Fails: this is the flat case, the two short sticks lying exactly along the long one.

(d) 2.5 + 2.5 = 5, and 4.9 < 5. Passes.

The two failures are certain, because they contradict the triangle inequality just proved. The two passes say only that the inequality rules nothing out. That such lengths really do make a triangle is the converse, which this lesson does not prove: see building a triangle from three sides.

Example 9

Limits on the third side

(a) Two sides of a triangle measure 8 and 11. Between what values must the third side x lie? How many whole-number values can it take?

(b) An isosceles triangle has sides of 4 and 9. Find its perimeter.

Solution

(a) Each side is shorter than the other two together, which gives three inequalities:

\begin{aligned} x &< 8 + 11 = 19 \\ 11 &< 8 + x \\ 8 &< 11 + x \end{aligned}

The second says x > 3. The third says x > -3, which adds nothing, since a length is positive. So 3 < x < 19. The whole numbers from 4 to 18 qualify, which is 15 of them. The ends are left out: x = 19 and x = 3 are both flat cases, since 8 + 11 = 19 and 3 + 8 = 11. In general, for two sides p and q the third side satisfies |p - q| < x < p + q.

(b) An isosceles triangle has two equal sides, and 4 and 9 differ, so one of them is repeated. If the sides are 4, 4, 9, then 4 + 4 = 8 is not more than 9, and there is no such triangle. If they are 4, 9, 9, then 9 < 4 + 9 and 4 < 9 + 9, so it works. The perimeter is 4 + 9 + 9 = 22.

Example 10

Unknown sides

For which values of x are x, x + 4 and 3x - 2 the sides of a triangle?

Solution

Each side must be shorter than the other two together. Write the three inequalities, one for each side, and simplify each.

For x: x < (x + 4) + (3x - 2) becomes x < 4x + 2, so \displaystyle x > -\frac{2}{3}.
For x + 4: x + 4 < x + (3x - 2) becomes x + 4 < 4x - 2, so x > 2.
For 3x - 2: 3x - 2 < x + (x + 4) becomes 3x - 2 < 2x + 4, so x < 6.

The sides must also be lengths, so positive. 3x - 2 > 0 gives \displaystyle x > \frac{2}{3}, and then the other two are positive as well. Of the three lower bounds, x > 2 is the strongest, so

2 < x < 6

Check the ends. At x = 2 the sides are 2, 6, 4, and 2 + 4 = 6 is flat. At x = 6 they are 6, 10, 16, and 6 + 10 = 16 is flat. In between, x = 4 gives 4, 8, 10, and 10 < 12.

Example 11

Two more forms of the inequality

A triangle has sides a, b, c and perimeter p = a + b + c. Show that (i) each side is less than half the perimeter, and (ii) the difference of any two sides is less than the third. Then answer: can a triangle with perimeter 30 have a side of 15.5? Can a triangle with sides 7 and 12 have a third side of 4?

Solution

(i) Start from c < a + b and add c to both sides, which keeps the inequality: 2c < a + b + c = p. So \displaystyle c < \frac{p}{2}. The argument is the same for a and b.

(ii) The inequalities a < b + c and b < a + c can be rewritten by subtracting b from the first and a from the second: a - b < c and b - a < c. One of a - b and b - a is the absolute value |a - b|, so |a - b| < c.

Now the questions. A perimeter of 30 means every side is less than 15, by (i), so a side of 15.5 is impossible. Sides 7 and 12 have |12 - 7| = 5, and by (ii) the third side must exceed 5, so a third side of 4 is impossible. Both answers are no.

Example 12

A second proof, through sides and angles in order

Prove AB < AC + CB again, this time using the fact that in a triangle the side opposite the larger of two angles is the longer side (sides and angles in order).

Solution

Given. \triangle ABC.

Prove. AB < AC + CB.

Plan. Lay the two short sides end to end along one line, and compare the far end with B.

Carry the line AC on past C to a point D with CD = CB, so C is between A and D.

Figure 6The second proof. CD = CB, so the angles at B and D of \triangle CBD are equal, and \angle ABD is bigger than \angle CBD.

CB = CD, so \triangle CBD is isosceles (it is a triangle because B is not on line AC), and the base angles theorem gives m\angle CBD = m\angle CDB. Because C is on the segment \overline{AD}, ray \overrightarrow{BC} is inside \angle ABD, and angle addition gives

m\angle ABD = m\angle ABC + m\angle CBD > m\angle CBD

Also \angle CDB is \angle ADB, since C is on ray \overrightarrow{DA}. Chaining the two: m\angle ABD > m\angle CBD = m\angle CDB = m\angle ADB.

In \triangle ABD, the angle at B is larger than the angle at D. The side opposite the angle at B is AD, and the side opposite the angle at D is AB, so the larger angle gives the longer side: AD > AB. By segment addition, AD = AC + CD = AC + CB. So AB < AC + CB.

Notice what this proof never did: it made no assumption. Everything it says is true of every triangle, and the inequality drops out of the last line, so it is a direct proof of the same fact. It is Euclid’s proof, Book I, Proposition 20, with the letters changed. It depends on the order of sides and angles, and that has its own proof, Euclid’s Propositions 18 and 19, which uses the exterior angle inequality and never mentions the triangle inequality, so nothing here runs in a circle. (The map sets the triangle inequality before sides and angles in order because the rule is met first, not because the order of sides and angles needs it for its proof.) The indirect proof in the procedure needed neither the order of sides and angles nor the exterior angle inequality, only base angles, angle addition, linear pairs and the angle sum.

Example 13

A point inside a triangle

P is a point inside \triangle ABC. Prove that PB + PC < AB + AC.

Solution

Given. P is inside \triangle ABC.

Prove. PB + PC < AB + AC.

Plan. The path B, P, C is tucked inside the path B, A, C, but no single triangle holds both paths. Carry BP on until it hits a side, and two triangles appear that between them do.

Carry ray \overrightarrow{BP} on until it leaves the triangle through the opposite side AC, at a point E. Since P is strictly inside, E is strictly between A and C, and P is strictly between B and E.

Figure 7Carrying BP on to E turns one inequality about a point inside a triangle into two triangle inequalities.

Apply the triangle inequality twice. In \triangle ABE (a triangle, since E is on AC and not at A, and B is not on AC):

BE < AB + AE \qquad (1)

In \triangle PEC (P is not on line AC, because it is inside):

PC < PE + EC \qquad (2)

Now build the chain. P is between B and E, so BP + PE = BE. Add BP to both sides of (2):

BP + PC < BP + PE + EC = BE + EC

By (1), BE + EC < AB + AE + EC, and E is between A and C, so AE + EC = AC. Together,

\begin{aligned} PB + PC &< BE + EC \\ &< AB + AE + EC = AB + AC \end{aligned}

Each < is strict because each comes from a triangle that really is a triangle. The inside path is always shorter than the path around it.

Boundaries

Indirect proof in general

The negation has to be complete. To prove “exactly one”, ruling out two leaves “none” open. To prove a strict “less than”, ruling out “more than” leaves “equal” open: the triangle inequality proof assumes “at least”, and the equal case, the one nearest to a real triangle, is where D = E.

“Not parallel” means “meet” only in a plane. In space, lines can miss each other without being parallel: skew lines. In space Example 4 is false. At a corner of a box, the upright edge and one bottom edge are both perpendicular to the other bottom edge, and they meet.

The contradiction must be with something already proved. Quoting a theorem whose proof used the statement being proved sends the argument in a circle. That is why Example 6 uses the exterior angle inequality and not the corresponding angles theorem (Exercise 35).

The sketch proves nothing. Figures of an assumption show things that cannot exist, and measuring them gives nonsense.

It shows that, not where. Example 2 proves the lines meet but not where, and Example 1 says two lines share at most one point, not that they share one.

The triangle inequality

Flat is the equality case. For three points on one line, AB \le AC + CB, with equality exactly when C is on the segment \overline{AB}. If C is between A and B, segment addition gives equality. If not, then A is between C and B, or B is between A and C, and then AC + CB is AB plus twice a positive length, so it is larger. The strict inequality belongs to real triangles, and the indirect proof reaches its contradiction because a triangle’s angle at C cannot be flat.

The proof needs C off the line AB twice. Once so that \triangle ACD and \triangle BCE are triangles with base angles, and once so that \angle ACB is less than 180^\circ. If C is on the line, AB = AC + CB can be true, and the proof rightly finds no contradiction: D, E and C are all one point.

The angle sum is used only in a weak form. It is used twice: to say that an angle of a triangle is below 180^\circ, and to count the angles of \triangle CDE in (2). If the angle sum were only known to be at most 180^\circ, which is all that holds without the parallel postulate, (2) would become “at least”, and the last line would still read m\angle ACB \ge 180^\circ. So this proof, like Euclid’s in Example 12, which uses no angle sum at all, never needs the postulate, and Euclid places the triangle inequality (Book I, Proposition 20) well before his first use of it (Proposition 29).

It proves one direction. The triangle inequality says which lengths are impossible. It does not say that every set of lengths that passes the test makes a triangle: that is the converse, with its own proof.

Where this goes next

The tangent radius theorem assumes a radius is not perpendicular to a tangent and finds a point of the tangent inside the circle. Example 6 is the engine of proving lines parallel. Example 1 has the shape of a uniqueness proof: suppose there are two, show they coincide. Outside geometry the same move shows that \displaystyle \sqrt{2} is irrational, and Example 5’s link to the contrapositive is taken further in proof by contraposition.

The converse of the triangle inequality, that any lengths with the longest less than the sum of the others make a triangle, is what lets you build one: building a triangle from three sides. The order of sides and angles used in Example 12 goes on into the hinge theorem, which compares two triangles that share two sides.

Using the triangle inequality again and again along a broken path gives a clean result: a path of any number of straight pieces from A to B is at least as long as the segment \overline{AB}. That is the precise sense in which a straight line is the shortest way between two points, and it is the reason the distance from a point to a line is measured along the perpendicular.

The same inequality turns up well outside triangles. Vectors obey |u + v| \le |u| + |v|, which is the triangle inequality drawn tip to tail (adding vectors tip to tail, the magnitude of a vector). Complex numbers obey it for their modulus, and real numbers for absolute value as distance: |x + y| \le |x| + |y|.

Exercises

Writing the assumption

Write the assumption that begins an indirect proof of each statement.

\ell \parallel m, for two different lines in one plane.
Every triangle has at most one obtuse angle.
m\angle X > 40^\circ.
Through the point P there is exactly one line perpendicular to \ell.

Measures that cannot be

Lines \ell and m are cut by t, and the corresponding angles \angle 3 and \angle 7 measure (5x - 12)^\circ and (3x + 20)^\circ, where x = 15. Prove that \ell \nparallel m.
Can a triangle have m\angle A = 92^\circ and m\angle B = 90^\circ? Answer with an indirect argument.
Rays \overrightarrow{EA}, \overrightarrow{EB}, \overrightarrow{EC} and \overrightarrow{ED} start at E, with E between C and D. m\angle AEC = (4x + 6)^\circ and m\angle BED = (6x - 20)^\circ, where x = 12. Prove that E is not between A and B.
A transversal t crosses \ell and m, and the two interior angles on the right of t measure 108^\circ and 75^\circ. Prove that \ell \nparallel m, and say on which side of t they meet.
For n = 18, can (3n + 5)^\circ, (4n - 2)^\circ and (2n + 13)^\circ be the angles of a triangle?

Writing indirect proofs

Prove that through a point P not on a line \ell there is at most one line perpendicular to \ell.
Prove that every triangle has at least two acute angles.
In a plane, \ell \perp t and m is a line that is not perpendicular to t. Prove that \ell and m meet.
\angle 1 and \angle 2 form a linear pair, and \angle 1 is obtuse. Prove that \angle 2 is acute.
Prove: if two lines are cut by a transversal and a pair of alternate interior angles are congruent, then the lines are parallel.

Can these be the sides of a triangle?

5, 8, 12
7, 7, 15
9, 10, 19
1.5, 2, 3.4
\displaystyle \frac{1}{2}, \frac{1}{3}, \frac{5}{6}

Limits on a side

Two sides of a triangle are 6 and 15. Between what two numbers must the third side lie?
An isosceles triangle has two sides of lengths 5 and 12. Find its perimeter.
For which values of x are x, x + 5 and 2x + 1 the sides of a triangle?
A triangle has perimeter 40. What is the largest whole-number length one of its sides can have?
Two sides of a triangle are 9 and 14, and the third is a whole number. How many different perimeters are possible?
Three towns A, B and C are joined by straight roads. A sign claims AC is 4 km, CB is 5 km and AB is 10 km. Can the sign be right?

The triangle inequality construction

In the indirect proof of the triangle inequality, AC = 8 and CB = 5. (a) If AB = 13, find AD and AE. What do you notice? (b) If AB = 15, find AE and DE.
\triangle ABC has m\angle A = m\angle B = 40^\circ. Build D and E as in Example 7. Find m\angle ACD and m\angle ECD, and check that m\angle ACB = 180^\circ - 2\,m\angle ECD.
A real triangle has AC = 6, CB = 4 and AB = 7. Build D and E as in the proof. Find AD, AE and ED. Which of D and E is nearer to A?

Using the triangle inequality

P is a point inside \triangle ABC. Prove that PA + PB + PC is more than half the perimeter of the triangle.
The diagonals AC and BD of a convex quadrilateral ABCD cross at O. Prove that AC + BD > AB + CD, and deduce that the two diagonals together are longer than half the perimeter.
M is the midpoint of side BC of \triangle ABC. Prove that \displaystyle AM < \frac{AB + AC}{2}. (Hint: carry AM on to a point A' with MA' = AM.)

Reasoning

Proving the converse of the corresponding angles theorem, a student writes: “Assume \ell \parallel m. Then corresponding angles \angle 2 and \angle 6 are congruent, which is what the Given says. So \ell \parallel m.” What is wrong?
Which steps of Example 4 use the plane?
Rewrite the proof in Example 5 without the word “assume”, using the contrapositive of the Vertical Angles Theorem. What changed?
A student shortens Example 6: “Assume \ell \nparallel m. Let \ell' be the parallel to m through P. By the corresponding angles theorem \ell' makes an angle congruent to \angle 2 with t, so \ell' is \ell.” In a course that proves the corresponding angles theorem from its converse, as Euclid and the converse lesson do, why does this prove nothing?
Figure 8 shows \triangle ABC drawn to size, with m\angle B = 70^\circ and m\angle C = 45^\circ; the dim ticks mark an assumption, AB = AC. An indirect proof that AB \neq AC would start from those ticks. What fact would it need to reach a contradiction? Finish the proof, and say what a course must have proved before the proof is allowed.

Figure 8The angles are drawn to size. The dim ticks are the assumption AB = AC.

Reasoning about the triangle inequality proof

The indirect proof of the triangle inequality assumes AB \ge AC + CB and not AB > AC + CB. Why is “at least” the right assumption?
Where exactly does that proof use the assumption? What changes if you drop it?
Follow the proof with AC = CB = 5 and AB = 10. Where do D and E land, and what does that say about C?
\triangle ABC has m\angle A = m\angle B = \alpha with \alpha < 60^\circ. Build D and E as in Example 7 and show that m\angle ECD = \alpha.
A student says: “A straight line is the shortest path between two points, so the triangle inequality needs no proof.” What is wrong with using that as the proof?
Where does the triangle inequality proof use that C is not on the line AB? What goes wrong, and what should go wrong, if it is?

Answers

\ell and m meet at a point.
Some triangle has at least two obtuse angles: “every” becomes “some”, “at most one” becomes “at least two” (negating a quantified statement).
m\angle X \le 40^\circ. The negation of > is \le, not <.
Either no line through P is perpendicular to \ell, or at least two are. Each case needs its own contradiction.
The angles measure 5(15) - 12 = 63^\circ and 3(15) + 20 = 65^\circ. Assume \ell \parallel m; the corresponding angles theorem then gives 63^\circ = 65^\circ, which is false. So \ell \nparallel m.
No. Assume it can. The angle sum gives m\angle C = 180^\circ - 92^\circ - 90^\circ = -2^\circ, but every angle measures more than 0^\circ. (Example 3 says so at once.)
The angles measure 4(12) + 6 = 54^\circ and 6(12) - 20 = 52^\circ. Assume E is between A and B; with E between C and D, \angle AEC and \angle BED are vertical angles, and the Vertical Angles Theorem gives 54^\circ = 52^\circ, which is false. So E is not between A and B.
The two add to 108^\circ + 75^\circ = 183^\circ. Assume \ell \parallel m; then same side interior angles add to 180^\circ, a contradiction. On the left of t the interior angles form linear pairs with these, so they measure 72^\circ and 105^\circ, adding to 177^\circ. That is less than 180^\circ, so by Euclid’s form of the parallel postulate the lines meet on the left.
No. The angles would be 59^\circ, 70^\circ and 49^\circ. If they belonged to a triangle they would add to 180^\circ, but they add to 178^\circ.
Assume two different lines m and n through P are perpendicular to \ell, at A and B. If A = B, then m and n share P and A, which differ since P is not on \ell, so m = n by Example 1, a contradiction. So A \neq B, and \triangle PAB has right angles at A and B, contradicting Example 3.
Assume some triangle has at most one acute angle. Then at least two of its angles measure 90^\circ or more, contradicting Example 3.
m \neq \ell, since \ell \perp t and m is not. Assume they do not meet; different lines in one plane, they are parallel. By the perpendicular transversal theorem, t is then perpendicular to m as well, contradicting the Given.
Assume \angle 2 is not acute, so m\angle 2 \ge 90^\circ. \angle 1 is obtuse, so m\angle 1 > 90^\circ, and then m\angle 1 + m\angle 2 > 180^\circ. But the Linear Pair Theorem gives m\angle 1 + m\angle 2 = 180^\circ, a contradiction. So \angle 2 is acute.
Say t crosses \ell at P and m at Q, P above Q, with alternate interior angles \angle a (at P, below \ell, right of t) and \angle b (at Q, above m, left of t) congruent. Assume \ell \nparallel m; as in Example 6 the lines meet at a point X off t, and \triangle PQX is a triangle. If X is right of t, \angle a is the triangle’s angle at P and \angle b is its exterior angle at Q, so m\angle b > m\angle a. If X is left of t, \angle b is the angle at Q and \angle a the exterior angle at P, so m\angle a > m\angle b. Either way the exterior angle inequality contradicts \angle a \cong \angle b. (Shorter: the angle vertical to \angle a corresponds to \angle b, so Example 6 applies.)
Yes. The longest side is 12, and 5 + 8 = 13 > 12.
No. 7 + 7 = 14, which is less than 15.
No. 9 + 10 = 19, which is not more than 19. It is the flat case.
Yes. 1.5 + 2 = 3.5 > 3.4.
No. \displaystyle \frac{1}{2} + \frac{1}{3} = \frac{5}{6}, which is not more than \displaystyle \frac{5}{6}. Flat again.
The third side is between 9 and 21. It must exceed 15 - 6 = 9 and be less than 15 + 6 = 21.
The perimeter is 29. Sides 5, 12, 12 work, since 12 < 5 + 12. Sides 5, 5, 12 do not: 5 + 5 = 10 < 12.
x > 2. The three inequalities are x < 3x + 6 (true for x > 0), x + 5 < 3x + 1, which gives x > 2, and 2x + 1 < 2x + 5 (always true). Sides must be positive, and x > 2 covers that. At x = 2 the sides are 2, 7, 5, which is flat.
19. Every side is less than half the perimeter, 20, so none is 20 or more. A side of 19 is possible: the sides 19, 11, 10 add to 40 and 19 < 21.
17. The third side x satisfies 5 < x < 23, so x = 6, 7, \dots, 22, which is 17 values, and each gives a different perimeter, 29 up to 45.
No. AC + CB = 4 + 5 = 9 is less than AB = 10, but for any three points AB \le AC + CB.
(a) AD = 8 and AE = 13 - 5 = 8, so D and E are the same point. The proof then gives m\angle ACB = 180^\circ: C is on AB, so there is no triangle. (b) AE = 15 - 5 = 10 and DE = 10 - 8 = 2. Now the proof gives m\angle ACB = 180^\circ + 2\,m\angle DCE, more than 180^\circ.
m\angle ACD = 70^\circ and m\angle ECD = 40^\circ. The base angles of \triangle ACD are \displaystyle \frac{180^\circ - 40^\circ}{2} = 70^\circ, and m\angle ACE = 100^\circ - 70^\circ = 30^\circ, so m\angle ECD = 70^\circ - 30^\circ = 40^\circ. Check: 180^\circ - 2(40^\circ) = 100^\circ, which is 180^\circ - 40^\circ - 40^\circ = m\angle ACB.
AD = 6, AE = 7 - 4 = 3 and ED = 6 - 3 = 3. E is nearer to A, the reverse of the assumed order, and they overlap by 3.
P is inside, so A, P, B form a triangle and PA + PB > AB. Likewise PB + PC > BC and PC + PA > CA. Adding: 2(PA + PB + PC) > AB + BC + CA, so PA + PB + PC is more than half the perimeter.
O is between A and C and between B and D. In \triangle AOB, AO + OB > AB. In \triangle COD, CO + OD > CD. Adding, and using AO + OC = AC and BO + OD = BD, gives AC + BD > AB + CD. The same argument in \triangle AOD and \triangle BOC gives AC + BD > AD + BC. Adding the two results, 2(AC + BD) > AB + BC + CD + DA, which is the perimeter, so AC + BD is more than half of it.
Carry AM on to A' with MA' = AM. BM = CM, and \angle AMB and \angle A'MC are vertical angles, so \triangle AMB \cong \triangle A'MC by SAS. Hence CA' = AB. In \triangle ACA', the triangle inequality gives AA' < AC + CA' = AC + AB, and AA' = 2AM. So 2AM < AB + AC, which is the claim.
The student assumed the Prove, not its negation, and reached a statement that agrees with the Given, which is no contradiction. An assumption that fits the facts is not shown true by that; the argument has the shape of affirming the consequent.
Two places. “Only one line is perpendicular to t at a point of t”, which gave P \neq Q in statement 4: in space a whole plane of lines is perpendicular to t there. And the link between “never meet” and “parallel”, which makes statement 5 the full negation and gives statement 12: in space, lines that never meet can be skew.
The Vertical Angles Theorem says: if \angle AEC and \angle BED are vertical angles, then \angle AEC \cong \angle BED. Its contrapositive: if \angle AEC \ncong \angle BED, they are not vertical angles. Here 70^\circ \neq 45^\circ, so they are not vertical. \overrightarrow{EC} and \overrightarrow{ED} are opposite rays, so it is \overrightarrow{EA} and \overrightarrow{EB} that are not, and E is not between A and B. The facts used are the same; nothing is assumed and then thrown away. Proving a conditional through its contrapositive in this way is a proof by contraposition.
The converse would rest on the theorem and the theorem on the converse, so neither would rest on the postulates. Two proofs leaning on each other show only that each follows from the other. Example 6 avoids this by using only the exterior angle inequality, which comes before both.
It needs a fact that turns equal sides into equal angles: the Base Angles Theorem, “if AB = AC, then \angle B \cong \angle C”, which with the ticks would give 70^\circ = 45^\circ. Assume, to reach a contradiction, that AB = AC. Then the angles opposite those sides, \angle C and \angle B, are congruent, so 45^\circ = 70^\circ, which contradicts the Given. So AB \neq AC. The Base Angles Theorem is proved from triangle congruence, so the proof is allowed only once congruence and the theorem are in place; before that, the assumption AB = AC leads to no contradiction anyone can cite. The indirect proof of the triangle inequality leans on the same theorem and waits for it too. A proof is only as far along as the facts already proved.
The Prove is “<”, and its opposite is “not <”, which is “\ge”, both “=” and “>”. Assuming only “>” would leave the equal case unexamined. The proof does cover it: equality puts D and E at the same point, and the argument allows D = E.
Only in placing the points: it makes E at least as far from A as D is, so the order is A, D, E, B. Every other step is true on any triangle. Drop it and the order may be A, E, D, B (Example 7). Then the counts give m\angle ACB = 180^\circ - 2\,m\angle ECD, which is below 180^\circ, and there is no contradiction.
AD = 5 and BE = 5 with AB = 10, so both D and E are the midpoint of \overline{AB}. Circles of radius 5 about A and B touch only at that midpoint, so C would have to be that same point. The “triangle” is flat, with a straight angle at C, 180^\circ. So this is the flat case, not a triangle.
Each of \triangle ACD and \triangle BCE has apex angle \alpha, so its base angles are (180^\circ - \alpha)/2 = 90^\circ - \alpha/2, and m\angle ACD + m\angle BCE = 180^\circ - \alpha. The angle at C is 180^\circ - 2\alpha. Since \alpha < 60^\circ, we have 90^\circ - \alpha/2 < 180^\circ - 2\alpha (this rearranges to 3\alpha/2 < 90^\circ), so D and E lie on the segment \overline{AB}. Also m\angle ACE = 90^\circ - 3\alpha/2, which is less than m\angle ACD = 90^\circ - \alpha/2, so E is between A and D, and the two base angles at C overlap along \angle ECD: m\angle ACD + m\angle BCE = m\angle ACB + m\angle ECD. That gives 180^\circ - \alpha = 180^\circ - 2\alpha + m\angle ECD, so m\angle ECD = \alpha.
“The straight line is the shortest path” is the triangle inequality in other words: the path A, C, B is at least as long as \overline{AB} is exactly the statement AB \le AC + CB. It restates the claim and gives no reason for it. The length of the bent path is the number AC + CB, and nothing already proved says it is at least AB; that is exactly what needs proof.
Twice. \triangle ACD and \triangle BCE must be genuine triangles for the base angles theorem to apply, and m\angle ACB < 180^\circ holds only because \angle ACB is an angle of a triangle. If C is on the line AB, the flat case AB = AC + CB can be true, and the proof should reach no contradiction. It doesn’t: D, E and C coincide, and \angle ACB is 180^\circ with nothing wrong.