Proving the triangle angle sum

When you reach for it

Once for each theorem, to earn it, and then every time after, to use it.

The triangle angle sum lesson states two theorems, uses them on numbers and gives the reason for each in prose: a line through one vertex, parallel to the opposite side, lays the three angles side by side along a straight line; and an angle outside a corner is whatever the inside angle there leaves of 180^\circ, which is exactly what the other two angles add to. Its exercises are the place to practice finding angles. Here those two reasons become two-column proofs, set up as in Given, prove and the diagram, with a named reason on every line as in giving a reason for every step. An added line gets a reason too, as every auxiliary line must.

On this map the order is fixed: the parallel postulate, then the Corresponding Angles Theorem, then the Alternate Interior Angles Theorem, then the Triangle Angle Sum Theorem, then the Exterior Angle Theorem, with the Exterior Angle Inequality a few lines behind it. Each may cite only what comes before it.

Once proved, each theorem is a one-line reason. Whenever a proof meets a triangle, “Triangle Angle Sum Theorem” turns its three angles into an equation. Whenever a triangle has a side running on past a corner, “Exterior Angle Theorem” trades the outside angle for two inside ones. Most of the worked examples are about writing those single lines so that they are justified, and about the three facts that follow from the angle sum and are quoted just as often.

The procedure

Proving the triangle angle sum

Triangle Angle Sum Theorem. The measures of the three angles of a triangle add to 180^\circ.

Given. \triangle ABC.

Prove. m\angle ABC + m\angle BAC + m\angle ACB = 180^\circ.

Plan. Nothing in a bare triangle puts its angles together, so add a line: through A, parallel to \overleftrightarrow{BC}. At A it makes a straight angle cut into three pieces. The middle piece is \angle BAC, and the outer two are alternate interior partners of \angle ABC and \angle ACB. Swap them in.

Proof 1.

The added line. Line 2 is where Figure 1’s dashed line comes from, and its reason is what makes drawing it legal. A is off \overleftrightarrow{BC} because A, B and C form a triangle, and through a point off a line at least one parallel exists: copy an alternate interior angle, and the parallel postulate lesson shows the copy never meets the line, without using the postulate. D and E are only names for points on either side of A. The line is given two conditions, through A and parallel, and nothing else; lines 7 and 8 have to prove what it does to the angles.

Where the postulate works. What the postulate adds is that only one parallel exists. That uniqueness is what the Corresponding Angles Theorem needs, and through it the Alternate Interior Angles Theorem. So line 2 does not lean on the postulate; lines 7 and 8 do.

The positions. Three lines lean on the diagram, which may supply positions but never sizes (what a diagram may assume). Line 5 needs \overrightarrow{AC} inside \angle BAE. Line 7 needs \angle DAB and \angle ABC to be alternate interior angles: between the parallels, on opposite sides of \overleftrightarrow{AB}, which holds because D and C are on opposite sides of it. Line 8 needs the same of E and B across \overleftrightarrow{AC}.

A second proof. Euclid’s own route, Book I, Proposition 32, adds two lines at C instead.

Proof 2.

Lines 6, 8 and 9 read positions from Figure 2, as Proof 1 did. In line 9, \angle ECD and \angle ABD sit in matching places, above \overleftrightarrow{BD} and on the D side of each crossing, and \angle ABD is \angle ABC because C is on \overrightarrow{BD}.

What each proof uses.

Both rest on the postulate, as any proof must, since the theorem fails where the postulate fails.

The same figure, a different target. Figure 2 is also the picture the triangle angle sum lesson uses for the angle outside the corner, where the two pieces at C show the exterior angle directly. Here they are aimed at the sum instead: line 5 brings in the inside angle at C, and line 12 ends on the three angles of the triangle. Euclid passes through the exterior angle on the way, since lines 6, 10 and 11 together already say m\angle ACD = m\angle BAC + m\angle ABC. This lesson states and proves that fact next, from the angle sum, in Proof 3.

Proving the exterior angle theorem

Extend one side of a triangle past a vertex. The angle between the extension and the other side at that vertex is an exterior angle; the triangle’s angles at the other two vertices are its remote interior angles.

Exterior Angle Theorem. An exterior angle of a triangle equals the sum of its two remote interior angles.

The set-up follows Given, prove and the diagram, with the angles numbered as in Figure 3.

Given. \triangle ABC; C is between B and D.

Prove. m\angle 4 = m\angle 1 + m\angle 2.

“Side \overline{BC} is extended past C to D” becomes a betweenness fact, because that is what the proof uses. No arcs appear: nothing is given equal.

Plan. Look at the straight angle at C. \angle 4 is whatever \angle 3 leaves of 180^\circ, since the two form a linear pair. The pair \angle 1, \angle 2 is also whatever \angle 3 leaves of 180^\circ, since the three angles of the triangle add to 180^\circ. Write both as 180^\circ - m\angle 3 and join them.

Proof 3.

Line 8 is there because of how substitution works: from a = b, it lets b replace a. Line 5 holds 180^\circ - m\angle 3, and line 7 has it on the right, so line 8 turns line 7 around first.

Three things about this proof are worth seeing.

The parallel lines enter at one line only. Line 6 cites the angle sum, which was proved with a line drawn parallel to a side, and so rests on the parallel postulate. No other line needs it.

Line 3 needs the extension to go past C. \overrightarrow{CB} and \overrightarrow{CD} are opposite rays only because C is between B and D. If D were between B and C, then \overrightarrow{CD} would be \overrightarrow{CB}, and \angle ACD would be \angle 3 itself (Exercise 6).

The theorem is about numbers. Line 9 equates measures; there is no single angle for \angle 4 to be congruent to. A proof that wants \cong at the end goes through the definition of congruent angles (Example 7).

The exterior angle inequality

A corollary is a result that follows from a theorem in a few lines. The Exterior Angle Inequality says an exterior angle of a triangle is greater than each of its remote interior angles.

Given. \triangle ABC; C is between B and D.

Prove. m\angle 4 > m\angle 1 and m\angle 4 > m\angle 2.

Plan. m\angle 4 is m\angle 1 plus something positive, and m\angle 2 plus something positive.

Proof 4.

The addition property of inequality says: if a > b, then a + c > b + c. Line 6 applies it to line 5, which gives m\angle 2 + m\angle 1 > 0^\circ + m\angle 1, and tidies each side. Line 4 turns line 3 around so that m\angle 4 may replace the sum, and substitution works inside an inequality just as inside an equation.

This proof of the inequality stands on the theorem, and so on the angle sum and the parallel postulate. Euclid proved the inequality first, as Proposition 16 of Book I, without the angle sum and before he used parallel lines at all (see Boundaries).

Citing the theorems

“Triangle Angle Sum Theorem” needs only that the three angles belong to one triangle. It gives an equation in measures, ready for substitution.

“Exterior Angle Theorem” and “Exterior Angle Inequality”, like any theorem, may be cited only when their if-part is on the page:

• a named triangle has one side extended past a vertex, so that the vertex is between the two ends of the longer segment, and
• the angles on the right-hand side are that triangle’s angles at its other two vertices.

Both are position facts, which the diagram may supply, as what a diagram may assume allows. In a figure with several triangles, name the triangle in the reason.

Worked examples

Boundaries

Only in the flat plane. Proofs 1 and 2 lean on the parallel postulate, through the Alternate Interior and Corresponding Angles Theorems, and Proof 3 leans on it at one line, line 6, through the angle sum. On a sphere, where the straightest lines are great circles, any two of them meet. So the parallel that line 2 of Proof 1 and line 3 of Proof 2 call for does not exist, a triangle’s angles add to more than 180^\circ, and in Proof 3 the line that fails is line 6. In the hyperbolic plane many lines through A miss \overleftrightarrow{BC}, so line 2 of Proof 1 stands, but no one of them makes both line 7 and line 8 true, and the angles add to less than 180^\circ.

The inequality asks for less. Euclid proves the Exterior Angle Inequality as Proposition 16 of Book I without the angle sum, using congruent triangles (which come later on this map, in the congruence criteria) and no parallel lines, so it holds in geometries where the parallel postulate fails. It still fails on a sphere, where lines close up into circles.

The order runs one way. On this map the angle sum comes first and the exterior angle theorem from it, so nothing proved from the angle sum may be used in its proof: not the exterior angle theorem (Example 10), not the polygon angle sum, not the three results of Examples 1 to 3. Euclid ran the other order, as the triangle angle sum lesson describes; either order is sound, but one proof may not mix them.

Two conditions for the added line. “Through A, parallel to \overleftrightarrow{BC}” fixes one line. “The line through A making \angle DAB \cong \angle ABC and \angle CAE \cong \angle ACB” asks for three conditions and assumes what lines 7 and 8 of Proof 1 prove.

Exterior to which triangle. The same angle can be exterior to one triangle and interior to another (Example 7), so the reason names the triangle. The side must run on past the vertex, and the inside angle at that vertex is never one of the two added.

Every route rests on something. A proof that turns the triangle a half turn about the midpoint of a side reaches the same straight angle, but only after showing that the turned copy lines up with the original, which is a fact about parallel lines again. A proof is as sound as the earliest facts it quietly uses.

Where this goes next

The angle sum feeds straight into the exterior angle theorem, as Proof 3 shows, the angle sum of a quadrilateral and the interior angle sum of any polygon; around a whole polygon the exterior angles add to 360^\circ, which is the exterior angle sum of a polygon. The Third Angle Theorem is why angle angle similarity needs only two angles, and why a congruence test with two angles and a side works wherever the side sits (the AAS theorem, in the congruence criteria). The right triangle corollary gives the 45^\circ and 30^\circ to 60^\circ angles of the special right triangles.

The Exterior Angle Inequality is the key step in proving that the larger angle lies opposite the longer side, and it drives indirect proofs such as Exercise 34. The Exterior Angle Theorem drives angle chases through figures made of several triangles (finding unknown angles), and in circles the proof of the inscribed angle theorem spots a central angle as an exterior angle of an isosceles triangle. Set out as boxes and arrows, Proof 1 or Proof 3 becomes a flowchart proof.

Exercises

Reading the proofs

1. Which lines of Proof 1 use the fact that \overleftrightarrow{DE} \parallel \overleftrightarrow{BC}? Why do the others not need it?
2. A student writes line 7 of Proof 1 as “\angle DAB \cong \angle ABC, alternate interior angles”. What is missing from the reason?
3. Rewrite the plan of Proof 1 with the parallel drawn through B instead of A. Which two angle pairs become alternate interior?
4. Line 9 of Proof 2 cites the Corresponding Angles Theorem. Extend \overleftrightarrow{EC} past C to a point F, and give a route to \angle ECD \cong \angle ABC that cites the Alternate Interior Angles Theorem instead.
5. Which line of Proof 3 depends on the parallel postulate, and through which theorem?
6. Line 3 of Proof 3 uses line 2, that C is between B and D. Suppose instead D is between B and C. What is \angle ACD then, and which line of the proof fails?
7. Write a different proof of the Exterior Angle Theorem. Keep lines 1 to 4 of Proof 3, and its line 6 as your line 5. Then set the two sums that equal 180^\circ equal to each other and subtract m\angle 3. Give a reason for each new line.
8. Why does Proof 4 need line 5? Can an exterior angle ever equal one of its remote interior angles?

Naming the reason

In Exercises 9 to 12, D is between B and C in Figure 6. Give the reason for each statement, and name the triangle where there is one.

9. m\angle ADB = m\angle C + m\angle DAC
10. m\angle ADC = m\angle B + m\angle BAD
11. m\angle ADC > m\angle BAD
12. m\angle ADB + m\angle ADC = 180^\circ

Using the theorems and the corollaries

13. In \triangle PQR and \triangle STU, \angle P \cong \angle S and \angle Q \cong \angle T. m\angle R = (4x - 9)^\circ and m\angle U = (2x + 25)^\circ. Find x and m\angle R.
14. Given. \triangle ABC with \angle C a right angle; m\angle A = (5y - 4)^\circ; m\angle B = (3y + 14)^\circ. Prove. y = 10, as a two-column proof that cites the Right Triangle Acute Angles Theorem. Then find both acute angles.
15. One angle of an equiangular triangle measures (3k + 6)^\circ. Find k.
16. In Figure 5, m\angle ACD = (3t + 5)^\circ and m\angle B = (5t - 13)^\circ. Find t, m\angle B and m\angle A.
17. In Figure 1, m\angle DAB = (2x + 10)^\circ, m\angle BAC = (3x + 6)^\circ and m\angle ACB = (4x - 16)^\circ. Find x and the three angles of the triangle.
18. In Figure 2, m\angle ECD = 48^\circ and m\angle ACE = 60^\circ. Find the three angles of \triangle ABC.
19. Use Figure 3. Given. C is between B and D; m\angle 1 = (2x + 9)^\circ, m\angle 2 = (3x - 4)^\circ and m\angle 4 = (6x - 15)^\circ. Prove. x = 20, as a two-column proof that cites the Exterior Angle Theorem. Then find all four angles.
20. In Figure 7, m\angle A = 64^\circ, m\angle ABD = 18^\circ and m\angle ACD = 25^\circ. Find m\angle BEC and m\angle BDC.
21. In Figure 8, m\angle A = 64^\circ and m\angle ABC = 70^\circ. Find m\angle BEC and m\angle ECD.
22. In Figure 6, \angle BAD \cong \angle C, m\angle B = 46^\circ and m\angle C = 58^\circ. Find m\angle BAC, m\angle DAC and m\angle ADB.
23. An exterior angle of a triangle measures (5x - 20)^\circ and one of its remote interior angles measures (3x + 10)^\circ. Use the Exterior Angle Inequality and the definition of angle measure to find every value x can take.

Writing proofs

Write each as a two-column proof.

24. Given. \triangle ABC with \angle C a right angle and \angle A \cong \angle B. Prove. m\angle A = 45^\circ.
25. Given. \triangle ABC and \triangle DEF with \angle C and \angle F right angles; \angle A \cong \angle D. Prove. \angle B \cong \angle E.
26. Given. In Figure 5, \angle ACB is a right angle and \overline{CD} \perp \overline{AB}. Prove. \angle DCB \cong \angle A.
27. Prove the Triangle Angle Sum Theorem with the line of Figure 1, citing the Alternate Interior Angles Theorem once and the Same Side Interior Angles Theorem once.
28. Given. \triangle ABC; C is between B and D; C is between A and E. Prove. m\angle BCE = m\angle A + m\angle ABC, using the Exterior Angle Theorem for \angle ACD and the Vertical Angles Theorem.
29. Given. D is between B and C on side \overline{BC} of \triangle ABC. Prove. m\angle ADB > m\angle C.
30. Given. \triangle ABC; B is between E and C; C is between B and D; \angle ABE \cong \angle ACD. Prove. \angle ABC \cong \angle ACB, using the Exterior Angle Theorem.

Reasoning

31. Suppose the line through A in Proof 1 is built by copying \angle ABC, so that \angle DAB \cong \angle ABC by construction. Which line still needs the parallel postulate, and why?
32. On a sphere a triangle can have three right angles. Which line of Proof 1 fails there?
33. Three positive angles add to 180^\circ. Does Proof 1 show that some triangle has exactly these angles? Explain.
34. Use the Exterior Angle Inequality to prove that a triangle cannot have two right angles.
35. In Example 9, explain why the bisector of the angle at B and the bisector of the exterior angle at C can never be parallel, whatever the triangle.