Geometry

Sides and angles in order

The three angles of a triangle are all different sizes. Without measuring, which side is the longest, and why can the answer never come out differently?

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Triangle inequalityMath 7

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Hinge theoremGeometry Indirect proofGeometry

Where you are 18 in Triangles

Geometry11

Triangles18

▴ 6 above Drawing triangles from given sides and angles: one, many or none SSS, SAS, ASA, AAS

Sides and angles in order

Hinge theorem Isosceles and equilateral triangles ▾ Show all 18 (7 more)

Why this exists

Two facts about triangles are already in hand. Equal sides lie opposite equal angles, the base angles theorem, and equal angles lie opposite equal sides, its converse. Between them they settle every question of the form “are these two equal?”

Most triangles have no equal sides at all, and for them those two theorems only say that the three angles are all different too. They do not say which angle is the big one. Given a triangle with angles 40^\circ, 60^\circ and 80^\circ, which side is longest? Given sides 7, 9 and 12, where is the widest corner?

The answer is as simple as it could be: sides and angles come in the same order. The longest side lies opposite the largest angle, the shortest side opposite the smallest, and the middle one opposite the middle one. This lesson states that exactly, in both directions, and proves it. Euclid builds both the triangle inequality and the hinge theorem on it.

Definition

The side opposite an angle of a triangle is the one side that does not touch that angle’s vertex (parts of a triangle). In \triangle ABC the side opposite \angle A is \overline{BC}, and its length is written a. In the same way b = CA is opposite \angle B and c = AB is opposite \angle C.

In words: within one triangle, the longer of two sides lies opposite the larger of the two angles, and the larger of two angles lies opposite the longer of the two sides.

In symbols, for any two sides of a triangle and the angles opposite them,

a > b \iff m\angle A > m\angle B

The double arrow \iff says that each side of it forces the other. Read it “exactly when”.

Intuition

Start from an isosceles triangle, the dashed one in Figure 1. Its two slanted sides are equal and so are its two base angles. Nothing is bigger than anything else, and the triangle is balanced.

Figure 1Moving the top vertex toward B opens the angle at B and stretches \overline{AC}, the side across from it.

Now slide the top vertex sideways toward B, keeping its height, to the solid position A. Two things happen at once. The vertex leans over B, so the angle at B opens and the angle at C closes. And the vertex moves away from C, so \overline{AC} gets longer and \overline{AB} gets shorter. The angle that opened is at B, and the side that stretched, \overline{AC}, is the one across the triangle from B. The two grow together.

In the solid triangle the angles are 70^\circ at B, 65^\circ at A and 45^\circ at C. The sides across from them come in the same order: \overline{AC} is longest, \overline{BC} is in the middle, and \overline{AB} is shortest. A picture shows one triangle, though, and the claim is about all of them. The proof below is what makes it certain.

Results

Theorem 1 (the longer side faces the larger angle). If two sides of a triangle are unequal, the angle opposite the longer side is larger than the angle opposite the shorter side. In \triangle ABC,

AC > AB \implies m\angle ABC > m\angle ACB

Theorem 2 (the larger angle faces the longer side). If two angles of a triangle are unequal, the side opposite the larger angle is longer than the side opposite the smaller angle. In \triangle ABC,

m\angle ABC > m\angle ACB \implies AC > AB

The equal case. Two sides of a triangle are equal exactly when the angles opposite them are equal. This is the base angles theorem together with its converse.

All three together. For any two sides a and b of one triangle and the angles A and B opposite them, exactly one line holds, and it holds in both directions:

\begin{aligned} a > b &\iff m\angle A > m\angle B \\ a = b &\iff m\angle A = m\angle B \\ a < b &\iff m\angle A < m\angle B \end{aligned}

So listing the three sides from shortest to longest lists the angles opposite them from smallest to largest. In particular the longest side is opposite the largest angle and the shortest side is opposite the smallest angle. Two sides tie exactly when their opposite angles tie. Combining the first two lines gives the forms with “at least”: a \ge b exactly when m\angle A \ge m\angle B.

Why it is true

Theorem 1. The idea is Euclid’s (Book I, Proposition 18): swing the shorter side down onto the longer one and see where it lands.

Given. \triangle ABC with AC > AB.

Prove. m\angle ABC > m\angle ACB.

Figure 2The proof of Theorem 1. AD = AB, so the two marked angles are equal. The one at B is part of \angle ABC, and the one at D is an exterior angle of the shaded triangle.

Statements	Reasons
1. There is a point D on \overline{AC} with AD = AB, strictly between A and C	AB is a positive length less than AC, so it can be marked off from A along \overline{AC} and falls short of C
2. m\angle ABD = m\angle ADB	Base angles theorem in \triangle ABD, where AB = AD. It is a real triangle because B is not on line AC
3. m\angle ABC = m\angle ABD + m\angle DBC	Angle addition postulate: D lies on \overline{AC} between A and C, so ray \overrightarrow{BD} is inside \angle ABC
4. m\angle ABC > m\angle ABD	From 3, since m\angle DBC > 0 (D \ne C)
5. m\angle ADB > m\angle DCB	Exterior angle inequality: \triangle DBC is a triangle (B is off line AC), and A, D, C lie on one line with D between, so \angle ADB is an exterior angle of it and \angle DCB is one of its remote interior angles
6. m\angle DCB = m\angle ACB	They are the same angle, since D lies on ray \overrightarrow{CA}
7. m\angle ABC > m\angle ACB	Chain 4, 2, 5 and 6

Written as one line, the chain is

m\angle ABC > m\angle ABD = m\angle ADB > m\angle ACB

The first step says the isosceles triangle’s angle at B is only part of the whole angle at B. The last step says its angle at D, being outside the shaded triangle BDC, beats the angle at C. The exterior angle inequality is the one fact that does real work.

Theorem 2. This is an indirect proof, and it is Euclid’s Proposition 19. It uses Theorem 1 and nothing new.

Given. \triangle ABC with m\angle ABC > m\angle ACB.

Prove. AC > AB.

Suppose not. Then AC \le AB, and since two lengths are either equal or one is smaller, there are two cases.

If AC = AB, the base angles theorem gives m\angle ABC = m\angle ACB. That contradicts the Given, which says the angle at B is larger.
If AC < AB, then Theorem 1 applies with the roles of B and C exchanged: the longer side \overline{AB} is opposite \angle ACB, so m\angle ACB > m\angle ABC. That contradicts the Given as well.

Both cases are impossible, so AC \le AB is false, and AC > AB.

The equal case is the base angles theorem and its converse, each proved in its own lesson. With it, the three lines of the summary follow: exactly one of a > b, a = b, a < b is true, the first matches m\angle A > m\angle B by Theorems 1 and 2, the second matches equality, and the third is the first with the letters swapped.

What the proof does not use. It never uses the triangle inequality. That matters, because Euclid proves the triangle inequality from Theorem 2 (Book I, Proposition 20), the route shown in Example 6 of proving the triangle inequality indirectly. If Theorem 1 needed the triangle inequality, that route would run in a circle. It does not: the only tools are marking off a length, the base angles theorem, angle addition and the exterior angle inequality.

Worked examples

Example 1

Ordering the sides from the angles

In \triangle XYZ, m\angle X = 48^\circ and m\angle Y = 75^\circ. List the sides from shortest to longest.

Solution

The order needs all three angles, and only two are given. The angles of a triangle add to 180^\circ (triangle angle sum), so

m\angle Z = 180^\circ - 48^\circ - 75^\circ = 57^\circ

In order, m\angle X < m\angle Z < m\angle Y, that is 48^\circ < 57^\circ < 75^\circ. Now read off the side opposite each angle, the one that does not touch its vertex: \overline{YZ} is opposite \angle X, \overline{XY} is opposite \angle Z, and \overline{XZ} is opposite \angle Y. The sides come in the same order as their angles:

YZ < XY < XZ

Example 2

Ordering the angles from the sides

In \triangle PQR, PQ = 7, QR = 12 and PR = 9. List the angles from smallest to largest.

Solution

First, these lengths do make a triangle, since the two shorter ones add to 7 + 9 = 16, which is more than 12. The sides in order are PQ < PR < QR. The angle opposite each side is at the vertex that side leaves out: \overline{PQ} is opposite \angle R, \overline{PR} is opposite \angle Q, and \overline{QR} is opposite \angle P. By Theorem 1 the angles follow the sides:

m\angle R < m\angle Q < m\angle P

The largest angle is at P, the one vertex not on the longest side. The theorem gives the order and nothing more: finding the actual sizes needs the law of cosines.

Example 3

Angles written in terms of x

In \triangle ABC, {m\angle A = (2x + 10)^\circ}, {m\angle B = (3x - 5)^\circ} and {m\angle C = (x + 25)^\circ}. List the sides from shortest to longest.

Solution

The angles add to 180^\circ:

\begin{aligned} (2x + 10) + (3x - 5) + (x + 25) &= 180 \\ 6x + 30 &= 180 \\ x &= 25 \end{aligned}

So {m\angle A = 2(25) + 10 = 60^\circ}, {m\angle B = 3(25) - 5 = 70^\circ} and {m\angle C = 25 + 25 = 50^\circ}, and these do add to 180^\circ. In order, m\angle C < m\angle A < m\angle B. The sides opposite are c = AB, a = BC and b = AC, so

AB < BC < AC

Notice that the expressions alone could not be ordered: 3x - 5 is less than x + 25 when x is small. Only the value of x that the angle sum forces decides it.

Example 4

Five segments, two triangles

In Figure 3, quadrilateral ABCD is cut by the diagonal \overline{BD}. In \triangle ABD, m\angle BAD = 50^\circ, m\angle ABD = 60^\circ and m\angle ADB = 70^\circ. In \triangle CBD, m\angle BCD = 80^\circ, m\angle CBD = 55^\circ and m\angle CDB = 45^\circ. List all five segments AB, BC, CD, DA and BD from shortest to longest.

Solution

Figure 3Two triangles share the side \overline{BD}, for Example 4.

The theorem compares sides of one triangle only, so work in each triangle separately and then look for a side they share.

In \triangle ABD: \overline{BD} is opposite the 50^\circ angle at A, \overline{AD} is opposite the 60^\circ angle at B, and \overline{AB} is opposite the 70^\circ angle at D. So

BD < AD < AB

In \triangle CBD: \overline{BC} is opposite the 45^\circ angle at D, \overline{CD} is opposite the 55^\circ angle at B, and \overline{BD} is opposite the 80^\circ angle at C. So

BC < CD < BD

The diagonal \overline{BD} is the longest side of the lower triangle and the shortest side of the upper one, so it links the two chains:

BC < CD < BD < AD < AB

The longest of the five is AB and the shortest is BC. The two chains joined into one only because the shared side is at the top of one chain and the bottom of the other. Otherwise some pair is left uncompared, as in Exercise 11.

Example 5

An indirect proof with the theorem

D is a point on side \overline{BC} of \triangle ABC, strictly between B and C. Prove that AD is shorter than at least one of AB and AC.

Solution

Figure 4A segment from a vertex to the opposite side, for Example 5.

Given. \triangle ABC, and D strictly between B and C.

Prove. AD < AB or AD < AC.

Suppose not. Then AD \ge AB and AD \ge AC together.

In \triangle ABD, the side \overline{AD} is opposite the angle at B and the side \overline{AB} is opposite the angle at D. Since AD \ge AB, the “at least” form of the theorem gives

m\angle ABD \ge m\angle ADB

In \triangle ADC, the side \overline{AD} is opposite the angle at C and \overline{AC} is opposite the angle at D. Since AD \ge AC,

m\angle ACD \ge m\angle ADC

Add the two:

m\angle ABD + m\angle ACD \ge m\angle ADB + m\angle ADC

D is between B and C, so \angle ADB and \angle ADC form a linear pair and the right side is 180^\circ. On the left, \angle ABD is \angle ABC and \angle ACD is \angle ACB, because D lies on both rays \overrightarrow{BC} and \overrightarrow{CB}. So

m\angle ABC + m\angle ACB \ge 180^\circ

But these are two angles of \triangle ABC, and the third angle is positive, so the angle sum makes them add to less than 180^\circ. The assumption has produced a contradiction, so it is false: AD is shorter than AB or shorter than AC. In other words, AD is shorter than the longer of the two.

A direct route exists too. The linear pair at D adds to 180^\circ, so one of its angles, say \angle ADB, is at least 90^\circ. The other two angles of \triangle ABD then add to at most 90^\circ, so m\angle ABD < 90^\circ \le m\angle ADB, and the side opposite the larger angle gives AB > AD. The indirect version is worth having because it needs no choice of which angle to look at.

Example 6

The perpendicular is the shortest segment

P is a point not on the line \ell, and F is the foot of the perpendicular from P to \ell. Q is any other point of \ell. Prove that PQ > PF. Deduce that in every right triangle the hypotenuse is the longest side.

Solution

Q \ne F and P is off the line, so P, F and Q form a triangle, with a right angle at F. Its angles add to 180^\circ, so the other two add to 90^\circ:

m\angle FPQ + m\angle PQF = 90^\circ

Both are positive, so each is less than 90^\circ, and in particular m\angle PQF < 90^\circ = m\angle PFQ. The side opposite the right angle at F is \overline{PQ}, and the side opposite the angle at Q is \overline{PF}. The larger angle faces the longer side, so

PQ > PF

For any right triangle, the same argument shows that the right angle is larger than each of the other two angles, so the side opposite it, the hypotenuse, is longer than each of the other two sides. This is a proof with no Pythagorean theorem in it. The lesson on the distance from a point to a line proves the first fact with Pythagoras and again with a mirror; here it is one application of Theorem 2.

Boundaries

One triangle. The theorem compares two sides of the same triangle. An 80^\circ angle in one triangle and a 40^\circ angle in another say nothing about the sides opposite them, because a triangle can be scaled up or down without changing any angle. Comparing sides of two different triangles needs more: two pairs of equal sides, in the hinge theorem.

Order, not proportion. The theorem ranks the sides; it does not measure them. In the triangle with angles 30^\circ, 60^\circ and 90^\circ the sides are 1, \displaystyle \sqrt{3} and 2 (30-60-90 triangles). The 60^\circ angle is twice the 30^\circ angle, but the side opposite it, \displaystyle \sqrt{3} \approx 1.73, is not twice 1. The exact link between a side and its opposite angle is the law of sines.

A real triangle. The proof needs B off the line AC, so that \triangle ABD is a triangle with base angles and \triangle DBC is a triangle with an exterior angle. Three points on one line break the statement. If C lies between A and B with CA = 1 and CB = 3, the “angles” at A and B are both 0^\circ, yet the sides opposite them, CB and CA, differ.

Strict gives strict, equal gives equal. A strict inequality between sides gives a strict one between the opposite angles, and the reverse. The “at least” form is not a separate theorem; it is the strict form and the equal case joined.

No parallel postulate. The proof uses the exterior angle inequality, not the angle sum. Euclid proves that inequality (Book I, Proposition 16) without parallels, and Propositions 18 and 19 come well before his first use of the parallel postulate in Proposition 29. The base angles theorem and angle addition need no parallels either, so the theorem holds wherever the exterior angle inequality does, the hyperbolic plane among them.

Where this goes next

The theorem is the bridge between angle facts and length facts, and several results cross it at once. Euclid’s proof of the triangle inequality lays the two short sides end to end and uses Theorem 2 in the triangle that appears; it is Example 6 of proving the triangle inequality indirectly. Example 6 above makes the perpendicular the shortest way to a line, the fact behind the distance from a point to a line.

The hinge theorem carries the comparison from one triangle to two: when two triangles have two pairs of equal sides, the larger included angle faces the longer third side. Euclid’s proof (Book I, Proposition 24) fits the two triangles together and finishes with Theorem 2; the hinge theorem lesson takes another road, through the triangle inequality.

In the special right triangles (30-60-90 and 45-45-90) the order of the sides is visible before a single length is computed, and in trigonometry it says the leg opposite the larger acute angle is the longer leg (opposite, adjacent and hypotenuse). When a triangle is named by its largest angle, as in classifying triangles by angles, this theorem is how you find that angle from the sides. The law of sines finally turns the order into exact numbers.

Exercises

Sides from angles

In \triangle ABC, m\angle A = 42^\circ, m\angle B = 81^\circ and m\angle C = 57^\circ. List the sides from shortest to longest.
In \triangle DEF, m\angle D = 118^\circ and m\angle E = 25^\circ. Name the longest side and the shortest side.
In \triangle GHK, m\angle G = m\angle H = 50^\circ. List the sides from shortest to longest.
\triangle LMN has a right angle at M, and m\angle L = 34^\circ. List the sides from shortest to longest.
The angles of \triangle PQR are in the ratio m\angle P : m\angle Q : m\angle R = 2 : 3 : 4. List the sides from shortest to longest.

Angles from sides

In \triangle ABC, AB = 8, BC = 5 and AC = 11. List the angles from smallest to largest.
In \triangle STU, ST = 6.5, TU = 6.2 and SU = 6.8. List the angles from smallest to largest.
In \triangle XYZ, XY = 2x + 1, YZ = 3x - 4 and XZ = x + 6, where x = 7. Which angle is largest, and which is smallest?
\triangle ABC has AB = AC = 9 and BC = 4. Compare its three angles.

Using the order

In \triangle ABC, m\angle A = 64^\circ and m\angle B = 58^\circ. Which side is longest? Are any two sides equal?
Quadrilateral ABCD has diagonal \overline{AC}. In \triangle ABC, m\angle BAC = 40^\circ and m\angle ABC = 95^\circ. In \triangle ACD, m\angle CAD = 70^\circ and m\angle ACD = 50^\circ. Order the sides of each triangle, and find the longest of the five segments AB, BC, CD, DA and AC. Which segment can this method not place in the order, and which comparisons does it leave open?
\triangle ABC has AB = 7, BC = 10 and CA = 15, and it is an obtuse triangle. Which angle is obtuse?
In \triangle ABC, m\angle A = (x + 20)^\circ, m\angle B = (2x)^\circ and m\angle C = (3x - 20)^\circ. Find x, and list the sides from shortest to longest.
In \triangle PQR, PQ = QR = 10 and m\angle Q = 50^\circ. Find the other two angles, and say whether PR is more or less than 10.
D is a point on side \overline{BC} of \triangle ABC, between B and C, with m\angle ADB = 100^\circ. Which is longer, AB or AD? Do these facts decide which of AC and AD is longer?

Reasoning

Prove that in a triangle with an obtuse angle, the side opposite the obtuse angle is the longest side.
The proof of Theorem 1 marks D on \overline{AC} with AD = AB. Where does it use AC > AB, and what becomes of the proof if AC = AB?
In \triangle ABC, AB > AC, and the bisector of \angle A meets \overline{BC} at D. Prove that \angle ADB is obtuse.
D is a point strictly between A and C on side \overline{AC} of \triangle ABC, with BD = BC. Prove that AB > BD.
The proof of Theorem 2 splits the assumption AC \le AB into two cases. Why does it need two, and why is there no third?
\triangle ABC has m\angle A = 80^\circ and \triangle DEF has m\angle D = 40^\circ. Give an example in which BC is shorter than EF, and explain why this does not contradict Theorem 2.
The sides of a triangle satisfy a < b < c. Prove that m\angle A < 60^\circ < m\angle C.

Answers

BC < AB < AC. \overline{BC} is opposite the 42^\circ angle at A, \overline{AB} is opposite the 57^\circ angle at C, and \overline{AC} is opposite the 81^\circ angle at B.
Longest \overline{EF}, shortest \overline{DF}. The third angle is m\angle F = 180^\circ - 118^\circ - 25^\circ = 37^\circ. The largest angle, at D, faces \overline{EF}, and the smallest, at E, faces \overline{DF}.
GK = HK < GH. m\angle K = 180^\circ - 50^\circ - 50^\circ = 80^\circ. \overline{GK} and \overline{HK} face the equal angles at H and G, so they are equal, and \overline{GH} faces the largest angle.
MN < LM < LN. m\angle N = 90^\circ - 34^\circ = 56^\circ. \overline{MN} faces 34^\circ, \overline{LM} faces 56^\circ, and the hypotenuse \overline{LN} faces 90^\circ.
QR < PR < PQ. With the angles 2k, 3k, 4k degrees, 9k = 180 gives k = 20, so the angles are 40^\circ, 60^\circ and 80^\circ. \overline{QR} faces \angle P, \overline{PR} faces \angle Q, and \overline{PQ} faces \angle R.
m\angle A < m\angle C < m\angle B. BC = 5 faces \angle A, AB = 8 faces \angle C, and AC = 11 faces \angle B.
m\angle S < m\angle U < m\angle T. TU = 6.2 faces \angle S, ST = 6.5 faces \angle U, and SU = 6.8 faces \angle T.
Largest \angle X, smallest \angle Y. With x = 7 the sides are XY = 15, YZ = 17 and XZ = 13 (a real triangle, since 13 + 15 > 17). The longest, \overline{YZ}, faces \angle X, and the shortest, \overline{XZ}, faces \angle Y.
m\angle B = m\angle C > m\angle A. \angle B and \angle C face the equal sides AC and AB, so they are equal. Each faces a side of 9, while \angle A faces the side of 4, so each is larger than \angle A.
\overline{BC} is longest, and AB = AC. The third angle is 180^\circ - 64^\circ - 58^\circ = 58^\circ, equal to m\angle B, so the sides opposite \angle B and \angle C are equal. \overline{BC} faces the largest angle, 64^\circ.
BC < AB < AC and AD < AC < CD, so \overline{CD} is longest. In \triangle ABC, m\angle BCA = 45^\circ, and BC, AB, AC face 40^\circ, 45^\circ, 95^\circ. In \triangle ACD, m\angle ADC = 60^\circ, and AD, AC, CD face 50^\circ, 60^\circ, 70^\circ. Joined at \overline{AC} the chains give BC < AB < AC < CD, but \overline{AD} is known only to be shorter than AC, so AD is left unplaced: it is never compared with AB or with BC. (Those lengths are fixed by the angles, and the law of sines gives BC \approx 0.65\,AC, AB \approx 0.71\,AC and AD \approx 0.88\,AC, but this theorem alone cannot say so.)
\angle B. An obtuse angle is the largest angle of its triangle, since the other two add to less than 90^\circ. The largest angle faces the longest side, CA = 15, which is opposite B.
x = 30, and BC < AC < AB. The angles add to 6x = 180, so they are 50^\circ, 60^\circ and 70^\circ at A, B, C. \overline{BC} faces \angle A, \overline{AC} faces \angle B, and \overline{AB} faces \angle C.
m\angle P = m\angle R = 65^\circ, and PR < 10. The equal sides make the base angles equal, each (180^\circ - 50^\circ) \div 2 = 65^\circ. \overline{PR} faces the 50^\circ angle and \overline{PQ} faces the 65^\circ angle at R, so PR < PQ = 10.
AB > AD; no. In \triangle ABD the 100^\circ angle is obtuse, so it is the largest, and \overline{AB} faces it. In \triangle ADC, m\angle ADC = 80^\circ faces \overline{AC}, while \overline{AD} faces \angle C, which is not given. If m\angle C = 30^\circ then AC > AD; if m\angle C = 90^\circ then AD > AC. Both fit the facts.
Let \angle C be obtuse. The angles add to 180^\circ, so m\angle A + m\angle B = 180^\circ - m\angle C < 90^\circ, and each of \angle A, \angle B is less than 90^\circ, so less than \angle C. By Theorem 2, c > a and c > b.
Only in step 1, to place D strictly between A and C. If AC = AB, then D = C: there is no triangle DBC and no exterior angle, and \angle ABD is all of \angle ABC. The chain shrinks to m\angle ABC = m\angle ACB, which is the base angles theorem, as it should be.
AB > AC gives m\angle C > m\angle B by Theorem 1. In \triangle ABD, m\angle ADB = 180^\circ - m\angle B - \tfrac{1}{2}m\angle A, and in \triangle ADC, m\angle ADC = 180^\circ - m\angle C - \tfrac{1}{2}m\angle A. So m\angle ADB > m\angle ADC. The two form a linear pair and add to 180^\circ, so the larger is more than 90^\circ.
\triangle BDC is isosceles, so m\angle BDC = m\angle BCD. Two equal angles of a triangle add to less than 180^\circ, so each is less than 90^\circ. Then m\angle ADB = 180^\circ - m\angle BDC > 90^\circ, since A, D, C lie on one line. In \triangle ABD, \overline{AB} faces this obtuse angle, so by Exercise 16 it is the longest side, and AB > BD.
The negation of AC > AB is AC \le AB, which means AC = AB or AC < AB. Each needs a different tool, the base angles theorem for the first and Theorem 1 for the second. There is no third case because two lengths are always equal or one is smaller.
Take \triangle ABC with angles 80^\circ, 50^\circ, 50^\circ and BC = 1, and \triangle DEF with angles 40^\circ, 70^\circ, 70^\circ and EF = 10. Then BC < EF although m\angle A > m\angle D. Theorem 2 compares two angles of one triangle; these are in different triangles, and scaling a triangle changes its sides without changing its angles.
By the theorem, m\angle A < m\angle B < m\angle C. So 3\,m\angle C > m\angle A + m\angle B + m\angle C = 180^\circ, giving m\angle C > 60^\circ, and 3\,m\angle A < 180^\circ, giving m\angle A < 60^\circ.