Hinge theorem

Given. \triangle ABC and \triangle DEF with AB = DE, AC = DF and m\angle BAC > m\angle EDF.

Prove. BC > EF.

Plan. Move a copy of \triangle DEF onto \triangle ABC so the two triangles share the side \overline{AB}. Then both third sides start at B, and a bisector and the triangle inequality compare them. This proof does not call on the order of sides and angles in a triangle (sides and angles in order) directly; it needs only SAS, the angle bisector and the triangle inequality (which is itself usually proved from that order).

Step 1, copy \triangle DEF onto \overline{AB}. On the same side of line AB as C, draw the ray from A that makes an angle of m\angle EDF with ray \overrightarrow{AB}. There is exactly one such ray (the protractor postulate). Since m\angle EDF < m\angle BAC and both rays are on the same side of AB, this ray lies inside \angle BAC. On it mark the point K with AK = DF.

Step 2, the copy is congruent. In \triangle ABK and \triangle DEF, AB = DE is given, \angle BAK \cong \angle EDF by construction, and AK = DF by construction. By SAS, \triangle ABK \cong \triangle DEF, so their corresponding sides give BK = EF (corresponding parts). The Prove has become

Step 3, bisect the angle between \overline{AK} and \overline{AC}. Now AK = DF = AC. The ray \overrightarrow{AK} is inside \angle BAC, so by the angle addition postulate

So ray \overrightarrow{AK} is not ray \overrightarrow{AC}, and K \ne C.

Draw the bisector of \angle KAC. The crossbar theorem says: if a point X lies in the interior of \angle BAC, then ray \overrightarrow{AX} meets segment \overline{BC} at a point strictly between B and C. The bisector runs inside \angle KAC, and \angle KAC lies inside \angle BAC, so the bisector’s points are in the interior of \angle BAC and the theorem applies. Call the crossing point M; it lies strictly between B and C.

Step 4, a second pair of congruent triangles. In \triangle AKM and \triangle ACM, AK = AC from Step 3, \angle KAM \cong \angle CAM because \overrightarrow{AM} bisects \angle KAC, and AM = AM. By SAS, \triangle AKM \cong \triangle ACM, so

Step 5, where is K? Nothing so far says where K falls. It can lie beyond \overline{BC} (Figure 3), inside \triangle ABC, or on \overline{BC} itself (Figure 4).

Case 1: K is not on line BC. This covers both K beyond \overline{BC} and K inside the triangle. M is on line BC and is not B, so line BM is line BC, and K is not on it. Then B, K and M are the corners of a real triangle, and the triangle inequality in \triangle BKM gives

The second line is Step 4, and the third is the segment addition postulate, since M is between B and C. The argument never asked which side of \overline{BC} the point K is on, so the inside and the beyond positions need no separate proofs.

Case 2: K is on line BC. Now \triangle BKM is flat and the triangle inequality cannot be cited, so this case needs its own line. The ray \overrightarrow{AK} is inside \angle BAC, so it crosses line BC at a point of the segment \overline{BC}, and it crosses that line only once, because two different lines meet at most once. So K is that point, on \overline{BC}. It is not C (Step 3), and it is not B, since m\angle BAK = m\angle EDF > 0^\circ. So K is strictly between B and C, and segment addition gives

since KC > 0.

Conclude. In every case BK < BC, and BK = EF, so BC > EF.

The hypothesis m\angle BAC > m\angle EDF was used twice: to put \overrightarrow{AK} inside \angle BAC, and to make \angle KAC a real angle with a bisector that lands strictly between B and C. The bisector AM is also the perpendicular bisector of \overline{KC}, because \triangle AKC is isosceles, so the proof can be read another way: B lies on the same side of that line as K, and every point on that side is nearer to K than to C (compare the perpendicular bisector theorem). That is how Petrunin’s book puts it.

The converse, by indirect proof.

Given. AB = DE, AC = DF and BC > EF.

Prove. m\angle A > m\angle D.

Assume the opposite, all of it: m\angle A \le m\angle D (indirect proof). There are two cases.

• If m\angle A = m\angle D, then \triangle ABC \cong \triangle DEF by SAS, so BC = EF. That contradicts BC > EF.
• If m\angle A < m\angle D, apply the hinge theorem with the triangles’ roles swapped: DE = AB, DF = AC and m\angle D > m\angle A, so EF > BC. That contradicts BC > EF too.

Both cases are impossible, so the assumption is false, and m\angle A > m\angle D.

Euclid proves the hinge theorem as Book I, Proposition 24, and the converse as Proposition 25 by this same two-case contradiction. His proof of Proposition 24 also fits the two triangles together, but finishes differently: it compares two angles of one triangle and uses sides and angles in order.

1. \overline{EF} is longer. The equal pairs meet at A and at D, so \angle A and \angle D are the included angles, and 83^\circ > 71^\circ gives EF > BC by the hinge theorem.
2. \overline{JL} is longer. The included angles are \angle K and \angle N, and 110^\circ > 104^\circ, so JL > MP.
3. \overline{RT} is longer. The equal pairs are \overline{RS} (shared) and \overline{ST}, \overline{SU}, meeting at S, so the included angles are \angle RST and \angle RSU, and 47^\circ > 39^\circ gives RT > RU.
4. At 100^\circ. Shoulder, elbow and hand make a triangle with sides 40 and 30 meeting at the elbow. The distance from shoulder to hand is the third side, and the wider elbow angle gives the longer third side.
5. BC = 10 and \displaystyle EF = 10\sqrt{2}. \triangle ABC is isosceles with apex angle 60^\circ, so its base angles are \tfrac{1}{2}(180^\circ - 60^\circ) = 60^\circ and it is equilateral. \triangle DEF is right-angled at D, so EF^2 = 10^2 + 10^2 = 200. Since 90^\circ > 60^\circ and \displaystyle 10\sqrt{2} > 10, the order agrees.
6. \angle B is larger. The equal pairs meet at B and at E, the third sides are AC = 18 and DF = 16, and 18 > 16 gives m\angle B > m\angle E by the converse.
7. \angle XYW is larger. The equal pairs are \overline{XY} (shared) and \overline{YZ}, \overline{YW}, meeting at Y. The third sides are XZ = 10 and XW = 11, so m\angle XYW > m\angle XYZ.
8. \angle CAD is larger. In \triangle ABD and \triangle ACD, AB = AC and AD = AD, with included angles \angle BAD and \angle CAD. The third sides are BD = 4 and DC = 6, so by the converse m\angle CAD > m\angle BAD.
9. 5 < x < 30. BC < EF gives m\angle A < m\angle D by the converse, so 3x - 15 < 75 and x < 30. An angle is more than 0^\circ, so 3x - 15 > 0 and x > 5.
10. 1 < x < 7. The hinge theorem gives EF < BC, so 2x + 1 < 15 and x < 7. The triangle inequality in \triangle DEF gives 10 < 7 + EF, so EF > 3, that is 2x + 1 > 3 and x > 1.
11. 2 < x < 12. The longer third side goes with the larger included angle, so 5x + 20 > 8x - 16, which gives x < 12. The second angle must be positive, 8x - 16 > 0, so x > 2. The first angle is then between 30^\circ and 80^\circ, a possible angle.
12. BC < AB = CA; the angles are m\angle A = 50^\circ, m\angle B = 65^\circ, m\angle C = 65^\circ. The third angle at P is 360^\circ - 130^\circ - 100^\circ = 130^\circ. \overline{BC} faces the 100^\circ angle and \overline{AB}, \overline{CA} face the two 130^\circ angles, so by the hinge theorem BC is shortest and by SAS AB = CA. The base angles are 25^\circ in \triangle APB, 40^\circ in \triangle BPC and 25^\circ in \triangle CPA, so {m\angle A = 25^\circ + 25^\circ}, {m\angle B = 25^\circ + 40^\circ} and {m\angle C = 40^\circ + 25^\circ}.
13. \overline{AB} is longer. \angle AMC forms a linear pair with \angle AMB, so m\angle AMC = 80^\circ. In \triangle AMB and \triangle AMC, AM = AM and MB = MC, and 100^\circ > 80^\circ gives AB > AC by the hinge theorem.
14. Obtuse. A right triangle with legs 7 and 8 has hypotenuse \displaystyle \sqrt{49 + 64} = \sqrt{113}, and 11^2 = 121 > 113, so the side 11 is longer and by the converse the angle opposite it is more than 90^\circ.
15. The longest side is 9, and 6^2 + 7^2 = 85 > 81 = 9^2, so as in Example 7 the angle opposite 9 is less than the right angle of a triangle with legs 6 and 7: it is acute. The largest angle of a triangle is opposite its longest side (sides and angles in order), so the other two angles are no larger and are acute too.
16. Twice: to put ray \overrightarrow{AK} inside \angle BAC in Step 1, and to make m\angle KAC > 0^\circ in Step 3, so that the bisector is a new ray landing strictly between B and C. If the angles were equal, \overrightarrow{AK} would be \overrightarrow{AC} and K would be C. There would be no angle KAC to bisect, and the triangles would simply be congruent, with BC = EF.
17. The only tool in Case 1 is the triangle inequality in \triangle BKM, which needs B, K, M to be a real triangle, that is K off line BM, which is line BC. Whether K is inside or beyond changes nothing in that inequality. When K is on BC the three points are on one line and the inequality cannot be cited, so segment addition is used instead.
18. The Prove is m\angle A > m\angle D, and its negation is m\angle A \le m\angle D, which contains both = and <. A contradiction in only one case would leave the other possible. SAS rules out the equal case (it forces BC = EF), and the hinge theorem with the triangles swapped rules out the smaller case (it forces EF > BC).
19. \overline{AC} is longer. In a parallelogram opposite sides are equal, so AB = DC, and consecutive angles add to 180^\circ, so m\angle DCB = 70^\circ. In \triangle ABC and \triangle DCB, AB = DC and BC = CB, with included angles \angle ABC and \angle DCB. Since 110^\circ > 70^\circ, the hinge theorem gives AC > DB.
20. If AB > AC, Example 6 shows \angle AMB is obtuse. For the other direction, suppose \angle AMB is obtuse. Its linear-pair partner \angle AMC is then 180^\circ minus more than 90^\circ, so it is acute, and m\angle AMB > m\angle AMC. In \triangle AMB and \triangle AMC, AM = AM and MB = MC, so the hinge theorem gives AB > AC.