Chapter 2

Triangles

Triangle properties, congruence and similarity criteria, the Pythagorean theorem, and key theorems.

Introduction

Triangle properties, congruence and similarity criteria, the Pythagorean theorem, and key theorems.

A triangle is a closed plane figure formed by three line segments connecting three non-collinear points. It is the simplest polygon and the basic building block of much of Euclidean geometry.

Every theorem about polygons, congruence, similarity, and area ultimately rests on properties of triangles.

2.1Classification of Triangles

Definition 2.1 (Classification of triangles). Triangles are classified by sides (equilateral, isosceles, scalene) or by angles (acute, right, obtuse).

Equilateral: all sides equal, all angles 60°.
Isosceles: at least two sides equal.
Scalene: all sides different.
Right: one 90° angle.
Acute: all angles < 90°.
Obtuse: one angle > 90°.

The two classifications are independent: a triangle carries one label from each list.

Intuition. An equilateral triangle is perfectly even, like a slice of pizza with equal sides. A right triangle has one perfect corner like the corner of a book.

The names tell you the "personality" of the triangle before you even draw it.

Example 2.2. Classify the triangle with sides $3, 4, 5$ .

Solution. By sides: all three lengths differ, so it is scalene. By angles: check the largest angle (opposite the longest side, $5$ ). Since $3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}$ , the converse of the Pythagorean theorem says that angle is exactly $9 0^{\circ}$ , so it is right.

A triangle with sides $5, 5, 6$ has two equal sides, so it is isosceles; it is not equilateral because the third side ( $6$ ) differs. Sanity check: the classifications by side and by angle are independent, so "scalene right" and "isosceles acute" are both perfectly valid combined labels.□

Exercises for §2.1 →

2.2Basic Properties and Angle Sum

Theorem 2.3 (Angle sum of a triangle). The interior angles of a triangle sum to

∠ A + ∠ B + ∠ C = 18 0^{\circ} .

This relies on the Euclidean parallel postulate; on a curved surface the sum can differ.

Proposition 2.4 (Longest side, largest angle). The longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.

Theorem 2.5 (Area of a triangle). From two sides $a, b$ and the included angle $C$ ,

Area = \frac{1}{2} ab sin C,

and from the three sides alone, by Heron's formula,

Area = s (s - a) (s - b) (s - c), s = \frac{a + b + c}{2},

where $s$ is the semi-perimeter.

Theorem 2.6 (Exterior angle theorem). An exterior angle equals the sum of the two remote (non-adjacent) interior angles.

Proof. The exterior angle and its adjacent interior angle form a straight line, so they sum to $18 0^{\circ}$ . The three interior angles also sum to $18 0^{\circ}$ . Subtracting the shared adjacent angle from both sums leaves the exterior angle equal to the other two interior angles.∎

Intuition. If you walk around the three corners of any triangle, you turn a total of 180° (a half-turn).

The biggest angle "faces" the longest side, like the biggest person sitting opposite the biggest slice.

Example 2.7. In $△ A BC$ with $∠ A = 5 0^{\circ}$ and $∠ B = 6 0^{\circ}$ , find the third angle and order the sides.

Solution. $∠ C = 18 0^{\circ} - 5 0^{\circ} - 6 0^{\circ} = 7 0^{\circ}$ . Step 2: Rank the angles: $∠ A (5 0^{\circ}) < ∠ B (6 0^{\circ}) < ∠ C (7 0^{\circ})$ . Step 3: Each side is opposite the angle of the same rank. Side $BC$ is opposite $∠ A$ , side $C A$ is opposite $∠ B$ , and side $A B$ is opposite $∠ C$ .

So $BC$ is the shortest side (opposite the smallest angle, $5 0^{\circ}$ ) and $A B$ is the longest (opposite the largest, $7 0^{\circ}$ ). Sanity check: the angles $5 0^{\circ} + 6 0^{\circ} + 7 0^{\circ} = 18 0^{\circ}$ . ✓□

Example 2.8 ((exterior angle):). In $△ A BC$ , $∠ A = 5 0^{\circ}$ and $∠ B = 6 0^{\circ}$ . The exterior angle at $C$ equals the two remote interior angles: $∠ A + ∠ B = 11 0^{\circ}$ . Check against the adjacent interior angle: interior $∠ C = 7 0^{\circ}$ , and $7 0^{\circ} + 11 0^{\circ} = 18 0^{\circ}$ . ✓

Exercises for §2.2 →

2.3Triangle Congruence and Similarity Preview

Definition 2.9 (Congruent triangles). Two triangles are congruent if corresponding sides and angles match exactly — same size and shape.

The main criteria (SSS, SAS, ASA, AAS, HL for right triangles) are proved using axioms and earlier propositions.

Similarity (same shape, possibly different size) is introduced later but relies on angle equality.

Intuition. Congruent triangles are identical twins; you can slide, flip, or turn one onto the other and they match perfectly.

Similar triangles are like family members: same angles, proportional sides, but one may be bigger.

Example 2.10. If △ABC has sides 3,4,5 and △DEF has sides 6,8,10, the triangles are similar (scale factor 2) but not congruent. All angles match because both are right triangles with the same acute angles.

Exercises for §2.3 →

Common Mistakes

Assuming any three lengths form a triangle: The triangle inequality must hold; 1+2 is not > 4.
Mixing up 'opposite' and 'adjacent' sides: The side opposite an angle does not touch that angle's vertex.
Thinking all isosceles triangles are equilateral: Only when the third side also equals the other two.
Forgetting the 180° sum when solving for an angle: Always subtract the two known angles from 180°.

Two geometric figures are congruent if one can be transformed into the other by a rigid motion (isometry): translation, rotation, or reflection. They have the same size and shape; corresponding parts are equal.

Congruence is the "same" relation for figures and is central to proofs in Euclidean geometry.

2.4Definition via Isometries

Definition 2.11 (Congruence). A congruence is a distance-preserving transformation (isometry) of the plane. If there exists an isometry mapping a figure $F_{1}$ onto a figure $F_{2}$ , then $F_{1} ≅ F_{2}$ .

For triangles, this reduces to the familiar SSS, SAS, ASA, AAS, HL criteria.

Intuition. Congruent figures are perfect copies; you can pick one up (without stretching or flipping in 3D if we disallow reflections sometimes) and lay it exactly on the other.

"Same shape and same size" is the everyday phrase.

Example 2.12. Two triangles with sides 6,8,10 are congruent (both are scaled 3-4-5 right triangles). You can rotate and translate one onto the other; all sides and angles match.

Exercises for §2.4 →

2.5Corresponding Parts

Notation (Order of vertices). When we write △ABC ≅ △DEF we mean angle A corresponds to angle D, side AB to DE, and so on. The order of vertices encodes the correspondence.

Theorem 2.13 (CPCTC). Corresponding parts of congruent triangles are congruent: once the triangles are known to be congruent, every pair of corresponding sides and every pair of corresponding angles are equal.

Intuition. The labels tell you which corner goes with which corner. If you wrote the vertices in the wrong order, the correspondence would be wrong even if the numbers match.

Once you know the triangles are congruent, every matching piece (sides, angles) must be equal.

Example 2.14. △ABC ≅ △DEF with correspondence A-D, B-E, C-F. Then ∠A = ∠D, AB = DE, BC = EF, etc. If we had written △ABC ≅ △DFE the correspondence would be different and the equalities would not hold.

Exercises for §2.5 →

2.6Criteria for Triangles

Theorem 2.15 (The congruence criteria). Two triangles are congruent as soon as any one of the following matches under a correspondence of vertices:

SSS — the three sides;
SAS — two sides and the angle included between them;
ASA — two angles and the side included between them;
AAS — two angles and a side not between them;
HL — in right triangles, the hypotenuse and one leg.

These five are shortcuts that guarantee an isometry exists without constructing it explicitly. Each criterion is proved from the axioms (often using superposition or auxiliary constructions).

Intuition. Instead of moving the whole triangle, we have quick "tests" using only 3 pieces of information that are enough to guarantee the triangles are identical twins.

HL is special for right triangles (hypotenuse-leg).

Example 2.16. Two right triangles each have a hypotenuse of $10$ and one leg of $6$ . By HL they are congruent. The reason: in a right triangle the third side (the other leg) is forced by Pythagoras, $1 0^{2} - 6^{2} = 100 - 36 = 64 = 8$ . So both triangles have sides $6, 8, 10$ , and now SSS finishes the job. Sanity check: HL is really "hypotenuse + one leg $\Rightarrow$ all three sides," which is why it works only for right triangles — the right angle is what lets Pythagoras recover the missing side. ✓

Exercises for §2.6 →

Common Mistakes

Writing the correspondence in the wrong vertex order: △ABC ≅ △DEF is not the same as △ABC ≅ △DFE.
Assuming SSS, SAS etc. work for any polygons: They are triangle-specific (for quadrilaterals you need more information).
Confusing congruence with similarity: Congruence requires equal size; similarity allows scaling.
Forgetting reflections: In some contexts (oriented congruence) reflections are disallowed; usually in school geometry reflections are allowed.

SSS (Side-Side-Side) is one of the fundamental congruence criteria: if three sides of one triangle are equal in length to the three sides of another triangle (corresponding sides), then the two triangles are congruent.

This criterion is powerful because it shows that three lengths completely determine the shape and size of a triangle (up to congruence).

2.7The SSS Criterion

Theorem 2.17 (SSS congruence). If in △ABC and △DEF we have AB = DE, BC = EF, and CA = FD, then △ABC ≅ △DEF, and consequently all corresponding angles are equal too.

The correspondence is determined by the order of the sides listed.

Intuition. If you have three sticks of lengths 3, 4, and 5, there is only one triangle (up to flipping) you can build with them. The angles are forced.

SSS is like a rigid lock: once the three sides snap together, the whole shape is fixed.

Example 2.18. Given △ABC with AB=5, BC=6, CA=7 and △DEF with DE=5, EF=6, FD=7. Then the triangles are congruent by SSS. In particular ∠A = ∠D, ∠B = ∠E, ∠C = ∠F, and the figures can be superimposed by rigid motion.

Exercises for §2.7 →

2.8Proof Sketch (Euclid style)

Euclid proves SSS (Prop. I.8) by a method of "application" and superposition, essentially showing that if sides match, the triangles can be made to coincide.

Proof. A modern proof uses the law of cosines, or constructs an auxiliary diagonal in the quadrilateral formed by the two triangles, shows the resulting triangles are congruent by SAS, and reads off the matching angles.∎

Intuition. Imagine two triangles made of rigid rods hinged at the vertices. If all three rods match in length, the two frames must fold into identical shapes.

The proof shows you can always move one onto the other without stretching or bending.

Example 2.19. Place △ABC and △DEF so that side BC coincides with EF (possible because BC=EF). Because AB=DE and AC=DF, the third vertices A and D must land on the same point (intersection of two circles), hence the triangles coincide.

Exercises for §2.8 →

2.9Why SSS Works and Uniqueness

Theorem 2.20 (Uniqueness from three sides). Given three side lengths satisfying the triangle inequality, there exists a triangle with those sides, and it is unique up to congruence.

This is why SSS is a congruence (not merely similarity) criterion: size is fixed by the actual lengths.

Intuition. Three lengths that can form a triangle "know" exactly what angles they must have. There is no freedom left.

This is different from SSS similarity, which would allow scaling.

Example 2.21. The set of all triangles with sides 3,4,5 is exactly the set of all right triangles with those leg/hypotenuse lengths; any two such are congruent to each other. You cannot have a "bigger" 3-4-5 triangle without scaling the sides.

Exercises for §2.9 →

Common Mistakes

Listing sides in wrong order: The correspondence must match the sides that are claimed equal; SSS requires the sides to correspond in the stated order.
Confusing with SSA (which is not always congruence): SSA (two sides and a non-included angle) can be ambiguous (the "ambiguous case").
Forgetting the triangle inequality check: 1,2,4 cannot form any triangle, so SSS does not apply.
Thinking SSS only works for equilateral: It works for any scalene triangle as well; the sides just have to match correspondingly.

SAS (Side-Angle-Side) congruence: if two sides and the included angle of one triangle are equal to two sides and the included angle of another, the triangles are congruent.

This is one of the most frequently used criteria because the included angle "locks" the shape between the two sides.

2.10The SAS Criterion

Theorem 2.22 (SAS congruence). If in △ABC and △DEF we have AB = DE, ∠B = ∠E, and BC = EF, then △ABC ≅ △DEF.

The angle must be the included angle, the one between the two sides.

Intuition. Think of two rods of fixed length with a hinge between them set at a fixed angle. The distance between the free ends is forced; the whole triangle is rigid.

The angle is the "glue" that fixes how the sides open.

Example 2.23. $△ A BC$ has $A B = 5$ , $∠ A BC = 6 0^{\circ}$ , $BC = 7$ . $△ D EF$ has $D E = 5$ , $∠ D EF = 6 0^{\circ}$ , $EF = 7$ .

Solution. The angle $∠ A BC$ is included between sides $A B$ and $BC$ ; likewise $∠ D EF$ sits between $D E$ and $EF$ . So this is genuinely SAS, not SSA. Step 2: The parts match: $A B = D E = 5$ , $∠ B = ∠ E = 6 0^{\circ}$ , $BC = EF = 7$ . Hence $△ A BC ≅ △ D EF$ .

We can verify the third side is forced by the law of cosines: $A C^{2} = 5^{2} + 7^{2} - 2 (5) (7) cos 6 0^{\circ} = 25 + 49 - 70 (0.5) = 74 - 35 = 39$ , so $A C = 39 \approx 6.24$ . The same computation gives $D F = 39$ , confirming $A C = D F$ . ✓□

Exercises for §2.10 →

2.11Proof Ideas

Proof. Euclid's argument (Prop. I.4) is superposition: place the equal angles on top of each other so their rays coincide. The first pair of equal sides makes one pair of endpoints coincide, the second pair makes the other coincide, so the third vertices land on the same point and the triangles coincide.∎

Modern treatments may instead use coordinates or the law of cosines to show the third side matches, then finish with SSS.

Intuition. Lay one triangle exactly on top of the other so the two sides and the angle between them match perfectly. Because the sides are the same length, the remaining vertex has nowhere else to go.

The proof shows the figures coincide completely.

Example 2.24. Place ∠B on ∠E so the rays coincide. Then point A lands on D because AB=DE. Point C lands on F because BC=EF. Therefore the segments AC and DF coincide and the triangles are identical.

Exercises for §2.11 →

2.12Included vs Non-included Angle

Pitfall. The angle must lie between the two given sides. If the angle is not included (SSA), the criterion fails in general and the ambiguous case can occur: zero, one, or two possible triangles.

Intuition. The angle has to be the "middle" one between the two sides you are given. If it is at the end, the side can swing and create different possible positions for the third vertex.

SSA is the "dangerous" case taught in trigonometry as the ambiguous case.

Example 2.25 ((included, unique):). Sides $10$ and $6$ with the $3 0^{\circ}$ angle between them (SAS). The third side is forced by the law of cosines: $x^{2} = 1 0^{2} + 6^{2} - 2 (10) (6) cos 3 0^{\circ} = 100 + 36 - 120 (0.866) \approx 136 - 103.9 = 32.1$ , so $x \approx 5.67$ . Exactly one triangle.

Example 2.26 ((non-included, ambiguous):). Now put the $3 0^{\circ}$ angle opposite the side of length $6$ , with the other given side $10$ (this is SSA). By the law of sines, $\frac{sin θ}{10} = \frac{sin 3 0 ^{\circ}}{6}$ , so $sin θ = \frac{10 \cdot 0.5}{6} = \frac{5}{6} \approx 0.833$ . This has two solutions in a triangle: $θ \approx 56. 4^{\circ}$ (acute) or $θ \approx 123. 6^{\circ}$ (obtuse), and both give valid triangles because in each case the remaining angle $18 0^{\circ} - 3 0^{\circ} - θ$ stays positive. So the same three data can build two non-congruent triangles — SSA is not a congruence criterion. Sanity check: had $sin θ$ come out greater than $1$ , there would be zero triangles; that is the other face of the ambiguous case.

Exercises for §2.12 →

Common Mistakes

Using SSA instead of SAS: Students often forget to check whether the given angle is included.
Assuming SAS works for any two sides and any angle: The angle must be between the sides.
Forgetting to verify correspondence: The equal angle must be between the pairs of equal sides in the correspondence.
Confusing with ASA: ASA uses two angles and the included side; SAS uses two sides and the included angle.

ASA (Angle-Side-Angle) congruence: if two angles and the included side of one triangle equal the corresponding two angles and included side of another, the triangles are congruent.

Because the third angles are automatically equal (180° sum), AAS is also a valid criterion, but the classical ASA emphasizes the included side.

2.13The ASA Criterion

Theorem 2.27 (ASA congruence). If in $△ A BC$ and $△ D EF$ we have $∠ A = ∠ D$ , $∠ B = ∠ E$ , and the included side $A B = D E$ (the side joining the two given angles), then $△ A BC ≅ △ D EF$ .

The word included is essential: the given side must be the one lying between the two given angles. Two angles plus that connecting side pin down both the shape (from the angles) and the scale (from the side length).

Intuition. If you know two corners and the exact length of the side between them, the whole triangle is forced. The other sides and angles have to adjust to make the angles fit.

The included side acts as the "base" that fixes the scale between the two angles.

Example 2.28. Suppose $△ A BC$ has $∠ B A C = 5 0^{\circ}$ , $∠ A BC = 6 0^{\circ}$ , and the side between them $A B = 6$ . Let $△ D EF$ have $∠ E D F = 5 0^{\circ}$ , $∠ D EF = 6 0^{\circ}$ , and $D E = 6$ .

Solution. The included side matches: $A B = D E = 6$ , and it is the side joining the two given angles in each triangle. Step 2: The two adjacent angles match: $∠ A = ∠ D$ , $∠ B = ∠ E$ . Step 3: By ASA, $△ A BC ≅ △ D EF$ .

As a consequence the third angles are forced equal: $∠ C = ∠ F = 18 0^{\circ} - 5 0^{\circ} - 6 0^{\circ} = 7 0^{\circ}$ , and the remaining sides match. Sanity check: the rays of the $5 0^{\circ}$ and $6 0^{\circ}$ angles rise from the two ends of the fixed base $A B = 6$ and can meet at only one point, so the third vertex — and hence the whole triangle — is uniquely determined. ✓□

Exercises for §2.13 →

2.14Relation to AAS and Third Angle

Since the sum of angles is $18 0^{\circ}$ , if two angles are known the third is determined.

Corollary 2.29 (AAS congruence). Two angles and a non-included side also determine the triangle uniquely up to congruence.

Intuition. Knowing two angles tells you the third for free. So you really only need two angles plus any side to fix the triangle (with proper correspondence).

The side can be opposite one of the angles; the law of sines then fixes the scale.

Example 2.30. △ABC: ∠A=40°, ∠B=70° ⇒ ∠C=70°. Side BC (opposite A) = 5. Then the triangle is determined. Another triangle with same angles and corresponding side 5 is congruent by AAS (or ASA after identifying the included side once the third angle is computed).

Exercises for §2.14 →

2.15Proof

Euclid proves ASA as Prop. I.26, along with AAS. The proof reduces to SAS, or argues by contradiction on the third side.

Proof. Suppose the sides adjacent to the base differ. Then two different triangles stand on the same base with the same base angles, and comparing them contradicts the exterior angle theorem (or the isosceles triangle theorem). So the sides agree, and SAS finishes the argument.∎

Intuition. Once the base and the two base angles are fixed, the arms of the angles must meet at a unique point (the third vertex).

If they met at two different points you would have two different third angles or sides, which contradicts the angle sum.

Example 2.31. Fix base BC. At B construct the given angle, at C construct the given angle. The two rays intersect at exactly one point A (because the angles determine unique directions). Hence unique triangle.

Exercises for §2.15 →

Common Mistakes

Confusing ASA with AAS: Both work, but students must be able to identify the included side correctly for ASA.
Assuming AAA is congruence: AAA gives similarity only; size is not fixed (you can scale the triangle and keep all angles).
Forgetting that the side must correspond correctly: The side must lie between the two given angles in the correspondence.
Using ASA when only one angle is given with two sides: That would be SSA or SAS depending on inclusion.

The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

It is one of the most famous and useful theorems in mathematics, with hundreds of proofs and applications in distance, vectors, and physics.

2.16Statement and Notation

Theorem 2.32 (Pythagorean theorem). For a right triangle $A BC$ with the right angle at $C$ ,

A B^{2} = A C^{2} + B C^{2},

where $A B$ , the side opposite the right angle, is the hypotenuse. Writing $c$ for the hypotenuse and $a, b$ for the legs,

c^{2} = a^{2} + b^{2} .

The hypothesis matters: this holds only for right triangles. For a general triangle the correct relation is the law of cosines, $c^{2} = a^{2} + b^{2} - 2 ab cos C$ , which reduces to the Pythagorean form exactly when $C = 9 0^{\circ}$ (so $cos C = 0$ ).

Intuition. If you build squares outward on each side of a right triangle, the two smaller squares together exactly cover the big square on the hypotenuse.

It is the "3-4-5" rule generalized: any right triangle obeys the same squaring relationship.

Example 2.33. A right triangle has legs $5$ and $12$ . The hypotenuse $c$ satisfies $c^{2} = 5^{2} + 1 2^{2} = 25 + 144 = 169$ , so $c = 169 = 13$ . This is the classic $5$ - $12$ - $13$ right triangle. Sanity check: the hypotenuse ( $13$ ) is longer than either leg, as it must be, and $1 3^{2} = 169$ recovers the sum. ✓

Exercises for §2.16 →

2.17Proofs (Euclid and Modern)

Euclid's proof (Prop. I.47) uses congruence of triangles obtained by dropping perpendiculars and rearranging areas. There are algebraic proofs (using similar triangles), vector proofs (dot product), and even proofs by rearrangement (van Schooten, etc.).

Proof. Drop an altitude from the right angle to the hypotenuse, creating two smaller right triangles similar to each other and to the original. Matching corresponding sides gives $a^{2} = c p$ and $b^{2} = c q$ , where $p$ and $q$ are the two segments the altitude cuts on the hypotenuse. Adding these and using $p + q = c$ yields

a^{2} + b^{2} = c (p + q) = c^{2} .

∎

Intuition. Euclid's proof is a giant jigsaw: he shows four copies of the triangle plus squares can be rearranged in two ways whose areas must match.

The similar-triangles proof says "the pieces are scaled versions of the big one, so their sizes multiply to give the squares."

Example 2.34. Using similar triangles: drop the altitude from the right angle to the hypotenuse, splitting it into segments $p$ and $q$ with $p + q = c$ . Each small triangle is similar to the whole (they share an acute angle plus the right angle, so AA applies). Matching corresponding sides gives $\frac{a}{c} = \frac{p}{a}$ and $\frac{b}{c} = \frac{q}{b}$ , i.e. $a^{2} = c p$ and $b^{2} = c q$ . Adding: $a^{2} + b^{2} = c (p + q) = c \cdot c = c^{2}$ . Sanity check with the $5$ - $12$ - $13$ triangle: $p = a^{2} / c = 25/13$ , $q = b^{2} / c = 144/13$ , and $p + q = 169/13 = 13 = c$ . ✓

Exercises for §2.17 →

2.18Converse and Applications

Theorem 2.35 (Converse of the Pythagorean theorem). If in a triangle $a^{2} + b^{2} = c^{2}$ , with $c$ the longest side, then the angle opposite $c$ is a right angle.

The converse lets you certify a right angle from side lengths alone, without measuring the corner.

Applications: distance formula in coordinate geometry is just Pythagorean on the differences $Δ x$ and $Δ y$ . In physics it appears in work, energy, and vector magnitude.

Intuition. If the sides satisfy the square relation, you know for sure there is a right angle even if you cannot see the corner.

GPS, video games, and construction all secretly use "square the differences and add" to find straight-line distances.

Example 2.36. Points $(1, 2)$ and $(4, 6)$ . $Δ x = 3$ , $Δ y = 4$ . Distance $= 9 + 16 = 5$ . This is Pythagorean applied to the horizontal and vertical legs of the right triangle formed by the points and the axes.

Exercises for §2.18 →

Common Mistakes

Applying it to non-right triangles: The relation fails for acute or obtuse triangles (use law of cosines instead).
Forgetting which side is squared alone: Only the hypotenuse (longest side, opposite right angle) goes on one side of the equation.
Confusing with area formula: (1/2)ab is area, not related directly to the squares in Pythagoras.
Assuming all 3-4-5 scaled triangles are the only right triangles: Any legs $a, b$ with hypotenuse $a^{2} + b^{2}$ work; $3$ - $4$ - $5$ is just the smallest integer example.

Two figures are similar if they have the same shape but not necessarily the same size. For triangles this means corresponding angles are equal and corresponding sides are proportional.

Similarity is the foundation of scale drawings, maps, fractals, and much of trigonometry.

2.19Definition for Triangles

Definition 2.37 (Similar triangles). △ABC ~ △DEF when $∠ A = ∠ D$ , $∠ B = ∠ E$ , $∠ C = ∠ F$ and the sides are in a common ratio

\frac{A B}{D E} = \frac{BC}{EF} = \frac{C A}{F D} = k,

the scale factor.

The order of vertices indicates the correspondence.

Intuition. Similar triangles are like photographs of the same scene taken from different distances: angles stay the same, but everything gets bigger or smaller together.

The sides "scale" by the same factor; the shape is identical.

Example 2.38. $△ A BC$ has angles $4 0^{\circ}, 6 0^{\circ}, 8 0^{\circ}$ . By the law of sines the sides are proportional to the sines of the opposite angles, giving sides of about $3, 4.04, 4.60$ . $△ D EF$ has the same three angles and sides about $6, 8.08, 9.19$ . Each side of $D EF$ is exactly twice the corresponding side of $A BC$ , so the scale factor is $k = 2$ and the triangles are similar. Sanity check: the angles are identical, so AA already guarantees similarity even before checking the side ratios. ✓

Exercises for §2.19 →

2.20AA, SAS, and SSS Similarity Criteria

Theorem 2.39 (Similarity criteria). Two triangles are similar as soon as any one of the following holds:

AA — two pairs of angles are equal (the third pair then matches, by the $18 0^{\circ}$ sum);
SAS similarity — two pairs of sides are proportional and the included angles are equal;
SSS similarity — all three pairs of sides are proportional.

Intuition. AA is the easiest: if two angles match, the third must, so shape matches.

SSS similarity says "all sides in the same proportions" — like a photocopy at a different zoom.

Example 2.40. Two triangles have sides 4,5,6 and 8,10,12. All ratios = 2, so SSS similarity; all angles equal even without measuring them.

Exercises for §2.20 →

2.21Applications

Similar triangles let us solve for unknown lengths using proportions. They justify the basic trigonometric ratios (sin, cos, tan are constant for a given angle across all similar right triangles). Used in shadow problems, map scaling, and perspective drawing.

Intuition. If you know a tree casts a 10 ft shadow when a 6 ft person casts a 4 ft shadow, the triangles are similar and you can set up 6/4 = h/10 to find the tree height h=15 ft.

Ratios replace measuring everything directly.

Example 2.41. A 1.7 m person stands 3 m from a lamppost and casts a 0.85 m shadow. The lamppost casts a 4.25 m shadow. Similar triangles: 1.7 / 0.85 = h / 4.25 ⇒ h = 8.5 m.

Exercises for §2.21 →

Common Mistakes

Confusing similarity with congruence: Similar triangles can be different sizes; congruent ones must be the same size.
Wrong correspondence: The angles that are equal must be matched correctly when writing the proportion.
Assuming all triangles with one equal angle are similar: Need two angles (or sides in proportion with angle).
Forgetting that similarity preserves angles but scales areas by $k^{2}$ : Linear dimensions scale by $k$ , areas by $k^{2}$ (and volumes, in 3D, by $k^{3}$ ).

The AA (Angle-Angle) similarity criterion states that if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.

Because the third angles are forced to be equal by the 180° sum, AA is the most commonly used similarity test for triangles.

2.22The AA Criterion

Theorem 2.42 (AA similarity). If $∠ A = ∠ D$ and $∠ B = ∠ E$ in △ABC and △DEF, then △ABC ~ △DEF: the third angles are equal too and the sides are proportional.

The correspondence is A↔D, B↔E, C↔F.

Intuition. If two corners of one triangle match two corners of another, the third corners must also match (they have to add up to 180°). The shapes are the same; only the sizes may differ.

You do not even need to look at the sides to know the angles are all the same.

Example 2.43. △ABC has ∠A=35°, ∠B=65° (so ∠C=80°). △DEF has ∠D=35°, ∠E=65°. Then the triangles are similar by AA; the third angles are both 80° and all corresponding sides will be in the same (unknown) ratio.

Exercises for §2.22 →

2.23Proof Using Parallel Lines or Basic Proportionality

Proof. Inside the larger triangle construct a line parallel to one side, cutting off a smaller triangle. The parallel makes corresponding angles along each transversal equal, so the cut-off triangle has the same angles as the given smaller triangle and is similar to it. The larger triangle is similar to the triangle it cuts off, so it is similar to the given one.∎

Another route uses the law of sines: $a / sin A$ equals the diameter of the circumcircle, so equal angles force proportional sides.

Intuition. Draw a line inside the bigger triangle parallel to the base; the small triangle on top has the same angles by alternate interior angles with the transversals.

The parallel forces the shapes to match.

Example 2.44. To prove AA: assume ∠A=∠D, ∠B=∠E. Construct on the larger triangle a line parallel to the base creating a small triangle with the vertex angle. The small triangle has the same angles as the given small one by parallels, hence is similar to it, and the original is similar to the small one by the same angles.

Exercises for §2.23 →

2.24Consequences and Basic Proportionality Theorem

Theorem 2.45 (Basic proportionality theorem (Thales)). A line parallel to one side of a triangle divides the other two sides proportionally. Conversely, a line dividing two sides proportionally is parallel to the third.

The theorem is a direct consequence of AA similarity.

Intuition. A line parallel to the base "cuts the sides in the same ratio." If it cuts one side in half, it cuts the other in half too.

This is why similar triangles give you easy proportion equations for unknown lengths.

Example 2.46. In △ABC, line DE || BC with D on AB, E on AC. Then AD/AB = AE/AC. If AD=3, AB=12, AE=4, then AC must be 16 to maintain the ratio 1/4.

Exercises for §2.24 →

Common Mistakes

Thinking you need all three angles given: Two suffice; the third is automatic.
Reversing the correspondence: Make sure the equal angles are matched to the correct vertices when writing the similarity statement.
Confusing AA similarity with AA congruence: There is no AA congruence; equal angles alone do not fix size.
Forgetting that sides must be corresponding: When setting up proportions, the sides opposite corresponding angles go together.

The isosceles triangle theorem (also called the base angles theorem) states that if two sides of a triangle are equal, then the angles opposite those sides are equal.

The converse is also true: if two angles are equal, the sides opposite them are equal.

2.25Statement and Converse

Theorem 2.47 (Isosceles triangle theorem). In △ABC, if $A B = A C$ then $∠ B = ∠ C$ : the base angles opposite the equal sides are equal. Conversely, if $∠ B = ∠ C$ then $A B = A C$ .

Intuition. If two sides are the same length, the "base" angles at the bottom must be the same; the triangle is "balanced."

If two angles are the same, the sides facing them must be the same length.

Example 2.48. $△ A BC$ has $A B = A C = 5$ and base $BC = 6$ . The theorem immediately gives $∠ B = ∠ C$ with no calculation, because $B$ and $C$ are the angles opposite the two equal sides $A C$ and $A B$ .

Solution.

If we do want their measure, drop the altitude from $A$ to the midpoint $M$ of $BC$ , giving a right triangle $A MB$ with $BM = 3$ and hypotenuse $A B = 5$ . Then $cos (∠ B) = \frac{BM}{A B} = \frac{3}{5} = 0.6$ , so $∠ B = ∠ C \approx 53.1 3^{\circ}$ . Sanity check: the apex angle is $18 0^{\circ} - 2 (53.1 3^{\circ}) \approx 73.7 4^{\circ}$ , and $53.1 3^{\circ} + 53.1 3^{\circ} + 73.7 4^{\circ} = 18 0^{\circ}$ . ✓

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Exercises for §2.25 →

2.26Proof

Proof. Construct the angle bisector from the apex $A$ , splitting △ABC into two triangles. They agree in the two sides $A B = A C$ , in the shared side along the bisector, and in the equal angles at $A$ , so they are congruent by SAS. Corresponding parts then give $∠ B = ∠ C$ .∎

This is Euclid's Prop. I.5, the pons asinorum; the altitude from $A$ serves just as well as the bisector.

Intuition. Draw the "middle line" from the top vertex straight down the middle of the base. This splits the triangle into two mirror-image right triangles that match by SAS or SSS.

Because the two small triangles are congruent, their bottom angles must be equal.

Example 2.49. Let M be the midpoint of base BC in isosceles △ABC with AB=AC. Then AM is also the angle bisector and altitude. △ABM and △ACM have AB=AC (given), BM=CM (midpoint), AM common. By SSS the triangles are congruent, so ∠ABM = ∠ACM.

Exercises for §2.26 →

2.27Applications and Equilateral Case

Corollary 2.50 (Equilateral triangles are equiangular). If all three sides are equal then all three angles are equal, and each is $6 0^{\circ}$ .

This is the isosceles triangle theorem applied three times, once for each pair of equal sides.

The theorem is used constantly in proofs involving symmetry and in constructing equilateral triangles.

Intuition. An equilateral triangle is a special isosceles where every side is a "base," so every angle must be equal — and they turn out to be 60°.

Many constructions and proofs "assume isosceles to get equal angles for free."

Example 2.51. Construct equilateral triangle on base BC: from B draw arc radius BC, from C draw arc radius CB; they intersect at A. Then AB=BC=CA by construction (radii), so all angles 60° by repeated application of the isosceles theorem.

Exercises for §2.27 →

Common Mistakes

Applying only to the base angles: The equal angles are the ones opposite the equal sides (the base angles if the equal sides are the legs).
Forgetting the converse: Equal angles imply equal opposite sides; this is useful for proving sides equal.
Assuming the altitude is always needed: The theorem holds without constructing anything; the proof may use construction but the statement does not.
Confusing with similarity: Isosceles triangles are not necessarily similar to each other unless angles match.

The triangle inequality states that the sum of any two sides of a triangle must be greater than the third side.

It is both a necessary and sufficient condition (with positivity) for three lengths to form a triangle, and it generalizes to the shortest-path property of straight lines in metric spaces.

2.28The Three Inequalities

Theorem 2.52 (Triangle inequality). For the sides $a, b, c$ of a triangle (all positive),

a + b > c, a + c > b, b + c > a .

All three must hold, though in practice you only need to check that the two shorter sides sum to more than the longest.

Equality ( $a + b = c$ ) holds only in the degenerate case when the "triangle" is flat — the three points are collinear and one lies between the other two, so the area is $0$ .

Intuition. You cannot have two short sides that together are shorter than or equal to the third; the ends would not meet to close the shape.

"The shortest distance between two points is a straight line" is the geometric reason: going via the third vertex would be longer.

Example 2.53. Sides $3, 4, 5$ : check all three sums against the remaining side. $3 + 4 = 7 > 5$ , $3 + 5 = 8 > 4$ , $4 + 5 = 9 > 3$ . All strict, so this is a valid (right) triangle.

Solution.

Sides $2, 3, 6$ : the two shortest give $2 + 3 = 5$ , which is not greater than $6$ . The inequality fails, so no triangle exists; the two short sides cannot reach across the long one and the figure collapses flat.

□

Example 2.54 ((finding the range):). For what lengths $x$ can $5, 7, x$ form a triangle? The three conditions are $5 + 7 > x$ , $5 + x > 7$ , and $7 + x > 5$ . These simplify to $x < 12$ , $x > 2$ , and $x > - 2$ (always true). Combining the binding ones: $2 < x < 12$ . So the smallest positive integer is $x = 3$ and the largest is $x = 11$ .

Exercises for §2.28 →

2.29Proof and Strict Inequality

Proof. The straight line is the shortest path. The two-leg route of length $a + b$ is a broken line joining the two ends of the side $c$ , so it is longer than the direct segment $c$ unless the break point lies on that segment, which is the degenerate case.∎

In vector terms the same statement reads

∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥,

with equality if and only if $u$ and $v$ are parallel and point in the same direction.

Intuition. Walking two sides of the triangle takes you from start to end, but the direct side is shorter or equal.

Only when you walk in a straight line (no bend) do the lengths add exactly.

Example 2.55. In the plane, place side $c$ on the $x$ -axis from $(0, 0)$ to $(c, 0)$ , and let the third vertex be $(x, y)$ . Then $a = x^{2} + y^{2}$ is the distance to $(0, 0)$ and $b = (x - c)^{2} + y^{2}$ is the distance to $(c, 0)$ . The two-leg path $a + b$ goes from $(0, 0)$ up to $(x, y)$ and back down to $(c, 0)$ , a bent path; the straight segment between the endpoints has length $c$ . Since the straight line is the shortest route, $a + b \geq c$ , and the bend is unavoidable whenever $y \neq = 0$ , giving the strict $a + b > c$ for any genuine (non-flat) triangle. Only when $y = 0$ and $0 \leq x \leq c$ does the path straighten out and $a + b = c$ . ✓

Exercises for §2.29 →

2.30Generalizations and Applications

In any metric space the distance satisfies

d (x, z) \leq d (x, y) + d (y, z)

for all points $x, y, z$ ; this is one of the defining axioms of a metric. In Euclidean geometry the inequality is strict unless the points are collinear with $y$ between $x$ and $z$ .

Used to prove the triangle inequality for vectors, norms, and in analysis (Minkowski inequality etc.).

In geometry it implies that the perimeter is greater than any side, useful in optimization and isoperimetric problems.

Intuition. "Going via a detour point is never shorter than going direct." This is true on maps, in graphs, and in curved spaces (with care).

GPS route planners use generalized versions when adding "via" points.

Example 2.56. In vector terms, $∥ A + B ∥ \leq ∥ A ∥ + ∥ B ∥$ . Take $A = (3, 0)$ and $B = (0, 4)$ . Then $A + B = (3, 4)$ , so $∥ A + B ∥ = 3^{2} + 4^{2} = 5$ , while $∥ A ∥ + ∥ B ∥ = 3 + 4 = 7$ . Indeed $5 < 7$ : the bend between the perpendicular vectors makes the direct hop shorter. Equality happens only when $A$ and $B$ point the same way; e.g. $B = 2 A = (6, 0)$ gives $∥ A + B ∥ = ∥(9, 0)∥ = 9$ and $∥ A ∥ + ∥ B ∥ = 3 + 6 = 9$ . ✓

Exercises for §2.30 →

Common Mistakes

Using $\geq$ instead of $>$ for actual triangles: The inequality is strict ( $>$ ) for non-degenerate (positive-area) triangles; the $\geq$ form allows the flat, collinear case.
Checking only one of the three inequalities: All three must hold.
Confusing with the Pythagorean relation: Pythagorean is equality for right triangles after squaring; triangle inequality is the linear version for any triangle.
Thinking it only applies to sides: The inequality applies to any three points in the plane: the distances between them must satisfy it.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter