Isosceles and equilateral triangles
Two sides of a triangle are equal. What does that force about its angles, does it work the other way round, and what if all three sides are equal?
Where you are 18 in Triangles
Why this exists
Three side lengths fix a triangle, but most of the time you are handed less, and the question is how much of the rest you can still work out without a protractor.
Here is a case that comes up constantly: two sides of a triangle have the same length. That is a fact about sides, yet it forces two of the three angles to be equal, and then one angle is enough to find the other two. Triangles with two equal sides are everywhere in geometry, often without being drawn as such: two radii of a circle, two sides of a regular polygon, the two rafters of a roof.
Very often the information arrives the other way round. Parallel lines, an angle bisector or a quick angle chase hand you two equal angles, and what you want is a length. The converse runs in that direction: it is how you prove a triangle is isosceles without measuring a single side, and how a surveyor finds the width of a river without crossing it.
Push both directions to a triangle whose three sides are all equal and every angle is known at once: mark three sides equal and you have been handed 60^\circ, 60^\circ and 60^\circ for nothing. This lesson proves all of it from the congruence criteria and the triangle angle sum.
Definition
A triangle is isosceles when at least two of its sides are congruent. The names of its parts come from those two sides.
- The two congruent sides are the legs.
- The third side is the base.
- The angle between the two legs, opposite the base, is the vertex angle.
- The two angles at the ends of the base are the base angles.
The word “base” has nothing to do with being at the bottom. The base is the side that is not a leg, wherever the triangle is drawn. In Figure 1, \overline{AB} \cong \overline{AC}, so the base is \overline{BC}, the vertex angle is \angle A, and the base angles are \angle B and \angle C.
Each base angle sits opposite one leg. The side opposite an angle is the one side that does not touch its vertex: \angle B is across the triangle from \overline{AC}, and \angle C is across from \overline{AB}. That is the reliable way to find the base angles in any drawing: find the two tick-marked sides, then look across from each.
Equilateral and equiangular
A triangle is equilateral when all three of its sides are congruent, and equiangular when all three of its angles are congruent. In \triangle ABC that reads
On a figure, three sides marked with the same number of ticks say “equilateral”, and three angles marked with the same number of arcs say “equiangular”, as in Figure 2.
An equilateral triangle is a special isosceles triangle. Isosceles asks for at least two congruent sides, and an equilateral triangle has three, so whichever side you call the base, the other two are congruent. (Sorting triangles by their sides and by their angles is the subject of classifying triangles.)
Intuition
Cut an isosceles triangle out of paper and fold it so that one leg lies exactly on the other. The crease runs from the vertex A to the middle of the base, and the two halves match.
The legs have the same length, so the fold carries B onto C, and the corner at B covers the corner at C exactly. The fold is a reflection, and rigid motions give one way to make this precise; this lesson uses congruent triangles instead.
Equal angles meet on the fold line
Now run the picture backwards. Stand at the two ends of a segment \overline{BC} and send a ray up from each end, both at the same angle to the segment. Fold the page along the line through the midpoint M that is perpendicular to \overline{BC}. B lands on C, and because the two angles are equal, the ray from B lands exactly on the ray from C. So the point where the rays cross folds onto itself: it sits on the fold line, and it is as far from B as from C.
Make the angles steeper and the rays meet higher up; make them shallower and they meet lower down. The meeting point moves up and down the fold line and never leaves it. The two sides from the base to that point are always a matched pair.
Three equal sides
Pick up an equilateral triangle, turn it a third of the way round, and put it back down. It fills the same outline, each corner sitting where the next one was. Nothing tells one corner from another: each sits between two sides of the same length, facing a third of the same length. With nothing to tell them apart, the three angles cannot differ, and three equal angles share the 180^\circ of the triangle angle sum equally, 60^\circ each.
A fold or a turn shows the result for the triangle in your hand. It does not show that the halves must match for every triangle, and “the crease hits the middle of the base” is itself something that needs a reason. The reasons come below.
Results
The Base Angles Theorem
Base Angles Theorem. If two sides of a triangle are congruent, then the angles opposite those sides are congruent.
In \triangle ABC:
In words: the base angles of an isosceles triangle are congruent.
Put this beside the triangle angle sum. Call the vertex angle v^\circ and each base angle b^\circ. The three angles add to 180^\circ, so
Two consequences follow at once.
- One angle gives all three. Know the vertex angle and each base angle is half of what is left. Know a base angle and the vertex angle is 180^\circ minus twice it.
- Base angles are acute. The vertex angle is more than 0^\circ, so 2b < 180 and b < 90. A right angle or an obtuse angle in an isosceles triangle can only be the vertex angle.
The converse of the Base Angles Theorem
Converse of the Base Angles Theorem. If two angles of a triangle are congruent, then the sides opposite those angles are congruent.
The triangle is then isosceles: the two given angles are its base angles, \overline{BC} is its base, and A is the vertex where the equal sides meet. The name says what it is: keep the setting (a triangle) and swap the hypothesis with the conclusion. As Converses and biconditional definitions explains, a converse needs a proof of its own.
Together with the Base Angles Theorem it makes one biconditional:
In words: a triangle is isosceles if and only if it has two congruent angles. Either fact may be used to get the other. And by the contrapositive of the Base Angles Theorem, a triangle whose angles are all different has sides that are all different.
Equilateral and equiangular
Three corollaries, results that follow from a theorem in a line or two, come from using the two theorems on more than one pair of sides.
Corollary 1. If a triangle is equilateral, then it is equiangular.
Corollary 2. If a triangle is equiangular, then it is equilateral.
Corollary 3. Each angle of an equiangular triangle measures 60^\circ.
Together they say three statements about a triangle are equivalent:
Two shortcuts follow. A triangle with two angles of 60^\circ is equilateral, because the third angle is 180^\circ - 60^\circ - 60^\circ = 60^\circ (Example 10 writes this out as a two-column proof). And an isosceles triangle with one angle of 60^\circ is equilateral, wherever that angle sits: if it is a base angle, the other base angle is 60^\circ too and the vertex angle is what is left, 60^\circ; if it is the vertex angle, the base angles share 120^\circ equally. Either way all three angles are 60^\circ.
Why it is true
The Base Angles Theorem: a triangle compared with itself
The proof, due to Pappus of Alexandria, compares the triangle with itself. One triangle can be named in two orders, and each order is a different correspondence, so a congruence statement can pair \triangle ABC with \triangle ACB.
Given. \overline{AB} \cong \overline{AC} in \triangle ABC.
Prove. \angle ABC \cong \angle ACB.
Plan. Match A \leftrightarrow A, B \leftrightarrow C, C \leftrightarrow B. Then side \overline{AB} matches \overline{AC}, \angle A matches itself, and \overline{AC} matches \overline{AB}: two sides and the angle between them, so the SAS postulate from SSS, SAS, ASA, AAS applies. Then the base angles are a pair of corresponding parts, and corresponding parts of congruent triangles are congruent.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \cong \overline{AC} | given |
| 2. \angle BAC \cong \angle CAB | reflexive property of congruence |
| 3. \overline{AC} \cong \overline{AB} | symmetric property of congruence: line 1 |
| 4. \triangle ABC \cong \triangle ACB | SAS postulate: lines 1, 2 and 3 |
| 5. \angle ABC \cong \angle ACB | corresponding parts of congruent triangles: line 4 |
Line 5 pairs the angle at the second letter of \triangle ABC with the angle at the second letter of \triangle ACB, which is the base angle at C. No extra line was drawn.
This is Proposition 5 of Book I of Euclid’s Elements, the pons asinorum or “bridge of asses”, long taken as the first hard crossing in the book. The more common modern proof draws the bisector of \angle A or the segment to the midpoint of the base (the crease in Figure 3) and matches the two halves; it is written out in proving the Base Angles Theorem.
The converse: the same comparison with ASA
The Base Angles Theorem cannot be quoted here: it assumes AB = AC, which is exactly what we want to prove. Pappus’s idea works again, with the ASA postulate in place of SAS.
Suppose \angle B \cong \angle C in \triangle ABC. Read the same triangle a second time in the order A, C, B, and set up the correspondence A \leftrightarrow A, B \leftrightarrow C, C \leftrightarrow B between \triangle ABC and \triangle ACB:
- \angle B in the first matches \angle C in the second, and \angle B \cong \angle C is given;
- side \overline{BC} matches side \overline{CB}, which is the same segment;
- \angle C in the first matches \angle B in the second, and \angle C \cong \angle B by the symmetric property of congruence.
Two angles and the side between them match, so by ASA, \triangle ABC \cong \triangle ACB. Corresponding parts of congruent triangles are congruent. Side \overline{AB} of the first triangle corresponds to side \overline{AC} of the second, so AB = AC.
A second route draws the bisector of \angle A, meeting \overline{BC} at D. Then \angle B \cong \angle C, \angle BAD \cong \angle CAD, and the shared side \overline{AD} is opposite \angle B and \angle C, so \triangle ABD \cong \triangle ACD by the AAS theorem from SSS, SAS, ASA, AAS, and again AB = AC. Euclid proves the converse as Proposition 6 of Book I, by contradiction.
Equilateral and equiangular: one theorem used twice
Both directions come from applying one isosceles theorem twice, with a different side as the base each time, and then joining the two results with the transitive property.
Corollary 1. Given. \triangle ABC is equilateral. Prove. \triangle ABC is equiangular.
| Statements | Reasons |
|---|---|
| 1. \triangle ABC is equilateral | given |
| 2. \overline{AB} \cong \overline{AC} | definition of equilateral: line 1 |
| 3. \angle C \cong \angle B | Base Angles Theorem: line 2 (\angle C is opposite \overline{AB}, \angle B is opposite \overline{AC}) |
| 4. \overline{BA} \cong \overline{BC} | definition of equilateral: line 1 |
| 5. \angle C \cong \angle A | Base Angles Theorem: line 4 (\angle C is opposite \overline{BA}, \angle A is opposite \overline{BC}) |
| 6. \angle B \cong \angle C | symmetric property of congruence: line 3 |
| 7. \angle B \cong \angle A | transitive property of congruence: lines 6 and 5 |
| 8. \triangle ABC is equiangular | definition of equiangular: lines 5, 6 and 7 |
The left panel of Figure 7 is lines 2 and 3; the right panel is lines 4 and 5. Twice is enough: the two uses share \angle C, so the transitive property gives the third pair free.
Corollary 2 runs the same way with the converse. From \angle A \cong \angle B it gives \overline{BC} \cong \overline{AC}, and from \angle B \cong \angle C it gives \overline{AC} \cong \overline{AB}. The transitive property joins them, and all three sides are congruent.
Corollary 3. Let each angle measure x^\circ. By the triangle angle sum,
Worked examples
The vertex angle is given
In Figure 8, \overline{AB} \cong \overline{AC} and m\angle A = 40^\circ. Find x and y.
The tick marks say \overline{AB} \cong \overline{AC}. The angles opposite those sides are \angle C and \angle B, so by the Base Angles Theorem they are equal: x = y.
The three angles add to 180^\circ, so
So x = y = 70^\circ.
Check. 40 + 70 + 70 = 180.
A base angle is given, and the triangle is turned
In Figure 9, \overline{PQ} \cong \overline{QR} and m\angle P = 52^\circ. Find m\angle R and m\angle Q.
Nothing here sits at the bottom, so read the ticks. The congruent sides are \overline{PQ} and \overline{QR}. The angle opposite \overline{QR} is \angle P, and the angle opposite \overline{PQ} is \angle R. So \angle P and \angle R are the base angles, and they are congruent:
The vertex angle is \angle Q, between the two legs. The angle sum gives
Base angles as expressions
In Figure 10, \overline{AB} \cong \overline{AC}, m\angle B = (3x + 8)^\circ and m\angle C = (5x - 16)^\circ. Find x and all three angles.
\angle B and \angle C are the base angles, since they are opposite the tick-marked sides. The theorem says they are equal, so set the expressions equal and solve:
Substitute into both expressions, since they must agree: 3(12) + 8 = 44 and 5(12) - 16 = 44. Each base angle measures 44^\circ, and the vertex angle is
The vertex angle and a base angle in terms of one unknown
In an isosceles triangle the vertex angle measures (2x + 10)^\circ and one base angle measures (x + 15)^\circ. Find x and all three angles.
SolutionThis time the theorem does not set the two expressions equal: one is the vertex angle, the other a base angle. What it gives is that the other base angle also measures (x + 15)^\circ. Now all three angles are known in terms of x, and they add to 180^\circ:
The vertex angle is 2(35) + 10 = 80 degrees and each base angle is 35 + 15 = 50 degrees.
Check. 80 + 50 + 50 = 180.
Equal angles give equal sides: find the third angle first
In \triangle DEF, m\angle D = 40^\circ and m\angle E = 70^\circ. The sides are DE = 4x - 7 and DF = 2x + 5. Find x and the length DE.
SolutionThe two given angles are different, so nothing is equal yet. Find the third angle:
Now \angle E \cong \angle F, and the converse of the Base Angles Theorem turns two equal angles into two equal sides: the sides opposite them. The side opposite \angle E is \overline{DF}, and the side opposite \angle F is \overline{DE}, so DE = DF:
Then DE = 4(6) - 7 = 17 and DF = 2(6) + 5 = 17. They agree, as they must. The third side, \overline{EF}, lies between the two equal angles; it is the base, and the theorem says nothing about its length.
An exterior angle
In Figure 11, D lies on \overrightarrow{BC} beyond C, m\angle ACD = 130^\circ, m\angle A = 65^\circ and AC = 9. Find BC.
\angle ACD is an exterior angle of \triangle ABC, so by the exterior angle theorem it equals the sum of the two far interior angles:
So \angle A \cong \angle B. The side opposite \angle A is \overline{BC} and the side opposite \angle B is \overline{AC}, so
Check. \angle ACB forms a linear pair with \angle ACD, so it is 50^\circ, and 65 + 65 + 50 = 180.
Measuring across a river
A tree stands at A on the far bank of a river. You stand at B on the near bank, directly across, so that \overline{AB} meets the bank at a right angle. You walk along the bank to C, stopping when the sight line to the tree makes a 45^\circ angle with the bank. You have walked 38\text{ m}. How wide is the river?
Take the banks to be straight and parallel, the tree to stand at the water’s edge, and every angle to be measured flat along level ground. Then A, B and C are the corners of a triangle, and the width of the river is AB, the perpendicular distance from the tree to the near bank.
In \triangle ABC, m\angle B = 90^\circ and m\angle C = 45^\circ, so
\angle A \cong \angle C. The side opposite \angle A is \overline{BC}, the path you walked, and the side opposite \angle C is \overline{AB}, the width of the river. So
Nothing on the far bank was measured. The two angles did all the work, and the one length you could reach stood in for the one you could not.
A bisector and a parallel
Given. In \triangle ABC, \overrightarrow{BD} bisects \angle ABC, with D on \overline{AC}; E is on \overline{AB} and \overline{DE} \parallel \overline{BC}.
Prove. EB = ED.
Plan. EB and ED are two sides of \triangle EBD, so it is enough to show that the angles opposite them, \angle EDB and \angle EBD, are congruent. The bisector ties \angle EBD to \angle DBC, and the parallel lines tie \angle DBC to \angle EDB.
| Statements | Reasons |
|---|---|
| 1. \overrightarrow{BD} bisects \angle ABC | given |
| 2. \angle EBD \cong \angle DBC | definition of angle bisector: line 1, with E on \overrightarrow{BA} |
| 3. \overline{DE} \parallel \overline{BC} | given |
| 4. \angle EDB \cong \angle DBC | Alternate Interior Angles Theorem: line 3, transversal \overleftrightarrow{BD} |
| 5. \angle DBC \cong \angle EDB | symmetric property of congruence: line 4 |
| 6. \angle EBD \cong \angle EDB | transitive property of congruence: lines 2 and 5 |
| 7. EB = ED | converse of the Base Angles Theorem: line 6, in \triangle EBD |
Line 7 names the triangle because the theorem is about two angles of one triangle. In \triangle EBD, \angle EDB is opposite \overline{EB} and \angle EBD is opposite \overline{ED}.
With numbers: if ED = 5 and AB = 12, then EB = 5 and AE = 12 - 5 = 7.
Equilateral: angles first, then sides
In \triangle JKL, m\angle J = (3x)^\circ, m\angle K = (2x + 20)^\circ and m\angle L = (x + 40)^\circ. Also JK = 4y - 1 and KL = 2y + 11. Find x, the three angles, y and the length of each side.
SolutionNothing yet says the triangle is special, so start with what every triangle has: its angles add to 180^\circ.
The angles are 3(20) = 60, 2(20) + 20 = 60 and 20 + 40 = 60 degrees. All three angles are equal, so \triangle JKL is equiangular, and therefore equilateral (Corollary 2). Its sides are equal:
Each side has length 4(6) - 1 = 23, and 2(6) + 11 = 23 agrees.
The order mattered. The side equation was only allowed once the angles had shown the triangle was equilateral; before that, JK and KL could have been any two lengths.
A proof from two 60^\circ angles
Given. In \triangle ABC, m\angle A = 60^\circ and m\angle B = 60^\circ.
Prove. \triangle ABC is equilateral.
Plan. The angle sum gives the third angle, which is also 60^\circ. Three equal measures make three congruent angles, and Corollary 2 turns equal angles into equal sides.
| Statements | Reasons |
|---|---|
| 1. m\angle A = 60 | given |
| 2. m\angle B = 60 | given |
| 3. m\angle A + m\angle B + m\angle C = 180 | triangle angle sum |
| 4. 60 + 60 + m\angle C = 180 | substitution: lines 1 and 2 into line 3 |
| 5. 120 + m\angle C = 180 | simplify: line 4 |
| 6. m\angle C = 60 | subtraction property of equality: line 5 |
| 7. 60 = m\angle B | symmetric property of equality: line 2 |
| 8. m\angle A = m\angle B | transitive property of equality: lines 1 and 7 |
| 9. 60 = m\angle C | symmetric property of equality: line 6 |
| 10. m\angle B = m\angle C | transitive property of equality: lines 2 and 9 |
| 11. \angle A \cong \angle B | definition of congruent angles: line 8 |
| 12. \angle B \cong \angle C | definition of congruent angles: line 10 |
| 13. \triangle ABC is equiangular | definition of equiangular: lines 11 and 12 |
| 14. \triangle ABC is equilateral | an equiangular triangle is equilateral (Corollary 2): line 13 |
Lines 7 and 9 exist only so that the transitive property has its equations the right way round: it needs a = b and b = c, with the shared quantity in the middle. Line 13 needs only two of the three pairs, since the transitive property supplies \angle A \cong \angle C.
Doubling a side of an equilateral triangle
In Figure 15, \triangle ABC is equilateral with side 6. Side \overline{BC} is continued past C to D so that CD = CA, and A is joined to D. Find m\angle ACD, m\angle CAD, m\angle BAD and the length BD.
\triangle ABC is equilateral, so each of its angles is 60^\circ (Corollaries 1 and 3). In particular m\angle ACB = 60^\circ.
\angle ACB and \angle ACD sit side by side along the straight line BD, so they form a linear pair:
In \triangle ACD, CD = CA, so it is isosceles with base \overline{AD}, and by the Base Angles Theorem \angle CAD \cong \angle CDA. Call each one x^\circ. The angle sum in \triangle ACD gives
So m\angle CAD = 30^\circ and m\angle CDA = 30^\circ. The ray \overrightarrow{AC} lies inside \angle BAD, so by the angle addition postulate
Finally BD = BC + CD = 6 + 6 = 12.
Check. \triangle ABD has angles 60^\circ at B, 90^\circ at A and 30^\circ at D, which add to 180^\circ. So it is a right triangle, and its longest side, BD = 12, is twice its shortest, AB = 6. Its third side AD, and the side ratios of every triangle with these angles, belong to special right triangles.
The triangle with a 36^\circ apex
In Figure 16, AB = AC, m\angle A = 36^\circ, \overrightarrow{BD} bisects \angle ABC with D on \overline{AC}, and BC = 4. Find m\angle BDC, BD and AD.
Start with the angles. AB = AC, so by the Base Angles Theorem \angle ABC \cong \angle C. The two share what is left of 180^\circ:
The bisector halves \angle ABC, so m\angle ABD = m\angle DBC = 36^\circ.
Now look for two equal angles inside each small triangle, and use the converse in each.
In \triangle BDC, the angle at D is
so \angle BDC \cong \angle C. The side opposite \angle BDC is \overline{BC} and the side opposite \angle C is \overline{BD}, so BD = BC = 4.
In \triangle ABD, m\angle A = 36^\circ = m\angle ABD. The side opposite \angle A is \overline{BD} and the side opposite \angle ABD is \overline{AD}, so AD = BD = 4.
So AD = BD = BC = 4, and the figure is really three isosceles triangles: \triangle ABC, \triangle BDC and \triangle ABD. The theorem was used once and its converse twice. The rest of \overline{AC}, the piece \overline{DC}, is not settled by equal angles alone; finding it needs similar triangles, a later topic.
Diagonals of a regular pentagon
ABCDE is a regular pentagon: its five sides are congruent and each of its angles measures 108^\circ (where the 108^\circ comes from is a later lesson, the polygon angle sum). The diagonals \overline{AC} and \overline{AD} split \angle A into three angles a, b and c, as in Figure 17. Find all three.
Look for triangles with two congruent sides. In \triangle ABC, \overline{AB} \cong \overline{BC} because they are sides of the regular pentagon. So \triangle ABC is isosceles with vertex angle \angle B = 108^\circ, and its base angles \angle BAC and \angle BCA are equal:
In the same way, \overline{AE} \cong \overline{ED}, so \triangle AED is isosceles with vertex angle \angle E = 108^\circ, and c = 36^\circ.
The diagonals lie inside \angle A, so the three angles fill it:
All three are 36^\circ: the two diagonals from one corner of a regular pentagon cut its angle into three equal parts, found from a figure that is not, at first sight, about isosceles triangles.
Boundaries
The theorem and its converse stand or fall separately. The Base Angles Theorem starts from two congruent sides; its converse starts from two congruent angles. Each needed its own proof, and the converse could not borrow the theorem, which assumes what the converse sets out to show (Converses and biconditional definitions).
The two angles must belong to one triangle. An angle of \triangle ABC congruent to an angle of \triangle DEF says nothing about any side. Example 8 names \triangle EBD for that reason.
The equal sides are the ones opposite the equal angles. They are not the sides that form the angles. The side between the two equal angles is the base, and neither theorem says anything about its length.
Neither theorem fixes the vertex angle’s size. It can be anything strictly between 0^\circ and 180^\circ; only the angle sum ties it to the base angles.
Equal is all they give. When two sides or two angles differ, both theorems are silent beyond “different”. Which side is longer, that the larger angle sits opposite the longer side, is a later result: the side-angle inequality.
The triangle must be a real one. Three points on one line make a flattened “triangle” whose angles at the two ends are both 0^\circ, and yet the two “sides” from the ends to the middle point can have any lengths. Euclid’s triangles are never flat, so he needs no warning; Petrunin, whose setting allows flat triangles, states the converse only for a triangle that is not degenerate.
Three sides, not two. Two congruent sides give two equal angles, not three. A triangle with angles 70^\circ, 70^\circ and 40^\circ is isosceles and not equilateral. One 60^\circ angle is not enough on its own either: 60^\circ, 45^\circ, 75^\circ has no two sides equal. It takes two 60^\circ angles, or two equal sides together with one 60^\circ angle.
Only triangles. With four or more sides, equal sides and equal angles come apart. A rhombus that is not a square has four equal sides and unequal angles; a rectangle that is not a square has four equal angles and unequal sides. A polygon with both properties has to be asked for both, which is why a regular polygon is defined as equilateral and equiangular.
Same shape, not same size. Every equiangular triangle has angles 60^\circ, 60^\circ, 60^\circ, yet a triangle of side 2 and a triangle of side 9 are not congruent. Matching all three angles fixes the shape and not the size, which is why AAA is not a congruence criterion.
The equalities do not need parallels; the numbers do. The proofs use only SAS and ASA. The formula v + 2b = 180 and the 60^\circ of Corollary 3 use the triangle angle sum, which rests on the parallel postulate. On a globe, a triangle with one corner at the North Pole and two on the equator a quarter of the way round from each other has three equal sides and three equal angles, but each angle is 90^\circ.
Where this goes next
Equal angles are easy to come by: parallel lines, bisectors and angle sums produce them all the time. The converse turns each such pair into a pair of equal lengths, and the theorem turns equal lengths back into angles, which is why the pair sits inside so many later proofs. The two-column proofs by the bisector and by the midpoint are practiced in proving the Base Angles Theorem, and matching a triangle with itself is the first of many proofs where the right correspondence does the work; Proving triangles congruent makes that a habit.
Circles are full of isosceles triangles, because any two radii are congruent. Join the center to two points of a circle and the base angles are equal, and that single step is how inscribed angles are measured. The same move gives the angles of a regular polygon, as in Example 13, and the base angles of an isosceles trapezoid and the equal angles of a kite.
In the special right triangles, the converse gives the equal legs of the 45-45-90 triangle at once, and cutting an equilateral triangle in half from a corner to the middle of the opposite side gives the 30-60-90 triangle, whose side ratios come from the Pythagorean theorem.
Euclid’s very first proposition is how to construct an equilateral triangle with compass and straightedge, and six of them fitted around one point make a regular hexagon, because 6 \times 60^\circ = 360^\circ. The same sum is why equilateral triangles cover a floor with no gaps, one of the three regular tessellations.
Exercises
Reading the theorems
- In \triangle XYZ, \overline{XY} \cong \overline{YZ}. Name the legs, the base, the vertex angle and the base angles.
- In \triangle JKL, \overline{JL} \cong \overline{KL}. Which angle is congruent to \angle J?
- In \triangle KLM, \angle K \cong \angle M and KL = 11. Which other side has length 11?
- In \triangle ABC, m\angle A = 30^\circ and m\angle B = 75^\circ. Which two sides are congruent?
Finding angles
An isosceles triangle has the angle given. Find its other two angles.
- Vertex angle 50^\circ.
- One base angle 38^\circ.
- An isosceles triangle has one angle of 100^\circ. Find the other two, and explain why there is only one possibility.
- An isosceles triangle has one angle of 70^\circ. Find every possible pair of other angles.
- In \triangle ABC, \overline{AB} \cong \overline{AC}. Side \overline{BC} is extended past C to a point D, and m\angle ACD = 128^\circ. Find the three angles of the triangle.
Finding sides
- In \triangle ABC, m\angle A = m\angle C = 38^\circ, AB = 12 and AC \approx 18.9. Find BC and m\angle B.
- An exterior angle at C of \triangle ABC measures 110^\circ, m\angle B = 55^\circ and AC = 8. Find BC and the three angles of the triangle.
- In \triangle DEF, m\angle F = 90^\circ, m\angle D = 45^\circ and DF = 6. Find EF.
- An equilateral triangle has a perimeter of 45 inches. How long is each side, and what does each angle measure?
- In \triangle PQR, m\angle P = 60^\circ, m\angle Q = 60^\circ and PQ = 10. Find QR, RP and the perimeter.
- In \triangle DEF, DE = DF = 8 and m\angle D = 60^\circ. Find m\angle E, m\angle F and EF.
Using algebra
- The base angles of an isosceles triangle measure (4x - 7)^\circ and (2x + 19)^\circ. Find x and all three angles.
- In an isosceles triangle the vertex angle measures (3y)^\circ and one base angle measures (y + 20)^\circ. Find y and all three angles.
- In \triangle ABC, \overline{AB} \cong \overline{AC}, and \overline{BC} is extended past C to D. If m\angle ACD = (5x + 10)^\circ and m\angle A = (3x - 20)^\circ, find x and the three angles of the triangle.
- The base angles of an isosceles triangle measure (x^2)^\circ and (3x + 40)^\circ. Find every value of x that works, and the vertex angle each one gives.
- The vertex angle of an isosceles triangle measures v^\circ. Write the measure of each base angle in terms of v.
- In \triangle JKL, m\angle J = 50^\circ, m\angle K = 65^\circ, JK = 2y + 1 and JL = 3y - 6. Find y and JK.
- In \triangle ABC, m\angle A = m\angle B = (4x)^\circ and m\angle C = (x + 18)^\circ. Also AC = 3y - 2 and BC = y + 10. Find x, the three angles, y and AC.
- One angle of an equilateral triangle measures (7x + 4)^\circ. Find x.
- An equilateral triangle has sides 4x - 3, 2x + 9 and 3y. Find x, y and the length of each side.
- The angles of \triangle PQR measure (x + 25)^\circ, (2x - 10)^\circ and (3x - 45)^\circ, and two of its sides measure 5y - 4 and 3y + 6. Find x, y and the length of each side.
More than one triangle
- In \triangle ABC, \overline{AB} \cong \overline{AC} and m\angle A = 20^\circ. Point D lies on \overline{AC} with \overline{BD} \cong \overline{BC}. Find m\angle DBC and m\angle ABD.
- \triangle ABD and \triangle CBD share the side \overline{BD}, with A and C on opposite sides of it. \overline{AB} \cong \overline{AD}, \overline{CB} \cong \overline{CD}, m\angle BAD = 50^\circ and m\angle BCD = 110^\circ. Find m\angle ABC and m\angle ADC.
- Points B, C and D lie on a line in that order. \overline{AB} \cong \overline{AC}, \overline{AC} \cong \overline{CD}, and m\angle BAC = 40^\circ. Find m\angle D and m\angle CAD.
- D is on side \overline{AC} of \triangle ABC. m\angle A = 30^\circ, m\angle ABD = 30^\circ, m\angle C = 60^\circ and AD = 5. Find BD, BC, DC and AC.
- In \triangle ABC, m\angle ABC = 80^\circ, m\angle A = 40^\circ, and \overrightarrow{BD} bisects \angle ABC with D on \overline{AC}. If BD = 7, find AD, m\angle C and m\angle BDC.
- Equilateral triangles \triangle ABC and \triangle ACD share the side \overline{AC} and lie on opposite sides of it. Find m\angle BAD and m\angle ABD.
- In equilateral \triangle ABC, the point D is on \overline{BC} and \overrightarrow{AD} bisects \angle BAC. Find m\angle BAD and m\angle ADB.
- ABCD is a square: four equal sides and four right angles. An equilateral triangle \triangle BCE is drawn on side \overline{BC}, outside the square. Find m\angle ABE and m\angle BAE.
- How many copies of one equilateral triangle can be placed around a single point, corner to corner, with no gaps and no overlaps? Explain with the angle measures.
Reasoning
- Explain why the base angles of an isosceles triangle are always acute, and why an equilateral triangle can never be a right triangle.
- A triangle has one angle of 60^\circ. Must it be equilateral? Explain, with an example if the answer is no.
- Explain why every equilateral triangle is isosceles, and why the converse is false.
- A student writes: “In \triangle ABC, \angle A \cong \angle B, so AB = BC.” What is wrong, and what should the conclusion be?
- Explain why the proof of the converse in “Why it is true” could not use the Base Angles Theorem in place of ASA.
- The proof of the Base Angles Theorem shows \triangle ABC \cong \triangle ACB, although only one triangle is drawn. Explain why this is a genuine comparison of two triangles, and which vertices it matches.
- In \triangle ABC, \overline{AB} \cong \overline{AC}. The base is extended past B to E and past C to F, so that E, B, C, F lie on one line. Prove that \angle ABE \cong \angle ACF.
- D is on \overline{AB} and E is on \overline{AC} in \triangle ABC, with \overline{DE} \parallel \overline{BC} and AD = AE. Prove that AB = AC.
- E is on \overrightarrow{BA} beyond A, so \angle CAE is an exterior angle of \triangle ABC. Ray \overrightarrow{AF} bisects \angle CAE and is parallel to \overline{BC}. Prove that AB = AC.
- In \triangle ABC, \overline{AB} \cong \overline{AC} and \angle A \cong \angle B. Prove that the triangle is equilateral.
Answers
- Legs \overline{XY} and \overline{YZ}; base \overline{XZ}; vertex angle \angle Y, between the legs; base angles \angle X and \angle Z, at the ends of the base.
- \angle K. \angle J is opposite \overline{KL} and \angle K is opposite \overline{JL}, so they are the two base angles.
- LM. \angle K is opposite \overline{LM} and \angle M is opposite \overline{KL}, so by the converse of the Base Angles Theorem LM = KL = 11.
- AB = AC. m\angle C = 180^\circ - 30^\circ - 75^\circ = 75^\circ, so \angle B \cong \angle C, and the sides opposite them are \overline{AC} and \overline{AB}.
- 65^\circ and 65^\circ. The base angles share 180^\circ - 50^\circ = 130^\circ equally.
- 38^\circ and 104^\circ. The other base angle is also 38^\circ, and the vertex angle is 180^\circ - 2(38^\circ) = 104^\circ.
- 40^\circ and 40^\circ. Base angles are acute, so the 100^\circ angle must be the vertex angle, and the base angles share 180^\circ - 100^\circ = 80^\circ.
- Two possibilities. If 70^\circ is the vertex angle, the others are 55^\circ and 55^\circ. If 70^\circ is a base angle, the others are 70^\circ and 180^\circ - 140^\circ = 40^\circ.
- m\angle A = 76^\circ, m\angle B = m\angle C = 52^\circ. \angle ACB forms a linear pair with \angle ACD, so it is 180^\circ - 128^\circ = 52^\circ; \angle B equals it as the other base angle; \angle A = 180^\circ - 104^\circ.
- BC = 12 and m\angle B = 104^\circ. \angle A is opposite \overline{BC} and \angle C is opposite \overline{AB}, so BC = AB. Then 180^\circ - 38^\circ - 38^\circ = 104^\circ. AC \approx 18.9 is the base and plays no part.
- BC = 8; the angles are m\angle A = 55^\circ, m\angle B = 55^\circ, m\angle C = 70^\circ. By the exterior angle theorem m\angle A + 55^\circ = 110^\circ, so m\angle A = 55^\circ, and \angle ACB is 180^\circ - 110^\circ = 70^\circ. \angle A \cong \angle B, so BC = AC.
- EF = 6. m\angle E = 180^\circ - 90^\circ - 45^\circ = 45^\circ, so \angle D \cong \angle E, and the sides opposite them, \overline{EF} and \overline{DF}, are equal.
- 15 inches, and each angle is 60^\circ. The three sides are equal, so 3s = 45 and s = 15; an equilateral triangle is equiangular, so each angle is 180^\circ \div 3.
- QR = 10, RP = 10, perimeter 30. m\angle R = 180^\circ - 60^\circ - 60^\circ = 60^\circ, so the triangle is equiangular and therefore equilateral, and every side equals PQ.
- m\angle E = m\angle F = 60^\circ and EF = 8. The base angles are equal and share 180^\circ - 60^\circ = 120^\circ, so each is 60^\circ. The triangle is equiangular, hence equilateral, so EF = DE = 8.
- x = 13; the angles are 45^\circ, 45^\circ and 90^\circ. The base angles are equal, so 4x - 7 = 2x + 19 and 2x = 26. Check: 4(13) - 7 = 45 = 2(13) + 19.
- y = 28; the angles are 84^\circ, 48^\circ and 48^\circ. The other base angle is also (y + 20)^\circ, so 3y + 2(y + 20) = 180 and 5y = 140. The two expressions are not set equal: one is the vertex angle and the other a base angle.
- x = 20; m\angle A = 40^\circ and m\angle B = m\angle C = 70^\circ. Each base angle is 180 - (5x + 10) = 170 - 5x, so (3x - 20) + 2(170 - 5x) = 180, which gives 7x = 140. Check: m\angle ACD = 110^\circ = 40^\circ + 70^\circ, as the exterior angle theorem says.
- x = 8, with vertex angle 52^\circ, or x = -5, with vertex angle 130^\circ. Setting the base angles equal, x^2 = 3x + 40, so (x - 8)(x + 5) = 0. For x = 8 each base angle is 64^\circ and the vertex angle is 180^\circ - 128^\circ; for x = -5 each is 25^\circ and the vertex angle is 180^\circ - 50^\circ. Both are genuine triangles.
- \displaystyle \left(\frac{180 - v}{2}\right)^\circ, which is \displaystyle \left(90 - \frac{v}{2}\right)^\circ. From v + 2b = 180, 2b = 180 - v.
- y = 7 and JK = 15. m\angle L = 180^\circ - 50^\circ - 65^\circ = 65^\circ, so \angle K \cong \angle L, and JL = JK: 3y - 6 = 2y + 1. Check: 3(7) - 6 = 15.
- x = 18; the angles are 72^\circ, 72^\circ, 36^\circ; y = 6 and AC = 16. From 4x + 4x + (x + 18) = 180, 9x = 162. \angle A \cong \angle B gives BC = AC, so y + 10 = 3y - 2 and 2y = 12. Check: 6 + 10 = 16.
- x = 8. Every angle of an equilateral triangle is 60^\circ, so 7x + 4 = 60 and 7x = 56.
- x = 6, y = 7, and each side is 21. From 4x - 3 = 2x + 9, 2x = 12; then 4(6) - 3 = 21, and 3y = 21 gives y = 7.
- x = 35, y = 5, and each side is 21. The angle sum gives 6x - 30 = 180, so x = 35 and the angles are 60^\circ, 60^\circ, 60^\circ. The triangle is equiangular, so equilateral: 5y - 4 = 3y + 6 gives y = 5, and 5(5) - 4 = 21.
- m\angle DBC = 20^\circ and m\angle ABD = 60^\circ. The base angles of \triangle ABC are (180^\circ - 20^\circ) \div 2 = 80^\circ. In \triangle BDC, \overline{BD} \cong \overline{BC}, so \angle BDC \cong \angle BCD, both 80^\circ, and m\angle DBC = 180^\circ - 160^\circ = 20^\circ. Then m\angle ABD = 80^\circ - 20^\circ.
- Both are 100^\circ. In \triangle ABD the base angles are (180^\circ - 50^\circ) \div 2 = 65^\circ; in \triangle CBD they are (180^\circ - 110^\circ) \div 2 = 35^\circ. At B and at D the two parts add: 65^\circ + 35^\circ = 100^\circ.
- m\angle D = m\angle CAD = 35^\circ. The base angles of \triangle ABC are (180^\circ - 40^\circ) \div 2 = 70^\circ. \angle ACD forms a linear pair with \angle ACB, so it is 110^\circ. \triangle ACD has \overline{AC} \cong \overline{CD}, so its base angles \angle CAD and \angle D are equal and share 180^\circ - 110^\circ = 70^\circ.
- BD = 5, BC = 5, DC = 5, AC = 10. In \triangle ABD, \angle A \cong \angle ABD, so BD = AD = 5. \angle BDC is exterior to \triangle ABD, so it is 30^\circ + 30^\circ = 60^\circ, and then m\angle DBC = 180^\circ - 60^\circ - 60^\circ = 60^\circ. In \triangle BDC, \angle BDC \cong \angle C gives BC = BD, and \angle DBC \cong \angle C gives DC = BD. So AC = AD + DC = 10.
- AD = 7, m\angle C = 60^\circ, m\angle BDC = 80^\circ. The bisector makes m\angle ABD = 40^\circ = m\angle A, so in \triangle ABD, AD = BD. Then m\angle C = 180^\circ - 80^\circ - 40^\circ = 60^\circ, and in \triangle BDC, m\angle BDC = 180^\circ - 40^\circ - 60^\circ = 80^\circ.
- m\angle BAD = 120^\circ and m\angle ABD = 30^\circ. \angle BAD is made of two 60^\circ angles. AB = AC = AD, so \triangle ABD is isosceles with vertex angle 120^\circ, and each base angle is (180^\circ - 120^\circ) \div 2 = 30^\circ.
- m\angle BAD = 30^\circ and m\angle ADB = 90^\circ. The bisector halves the 60^\circ angle at A. In \triangle ABD, m\angle B = 60^\circ, so m\angle ADB = 180^\circ - 60^\circ - 30^\circ = 90^\circ.
- m\angle ABE = 150^\circ and m\angle BAE = 15^\circ. m\angle ABE = 90^\circ + 60^\circ. AB = BC = BE, so \triangle ABE is isosceles with vertex angle 150^\circ, and each base angle is (180^\circ - 150^\circ) \div 2 = 15^\circ.
- Six. Each corner is 60^\circ and the angles around a point make 360^\circ, so 360^\circ \div 60^\circ = 6 fit exactly.
- The two base angles are equal, say b^\circ each, and with the vertex angle they add to 180^\circ. The vertex angle is more than 0^\circ, so 2b < 180 and b < 90. Every angle of an equilateral triangle is 60^\circ, which is not 90^\circ, so it has no right angle (and no obtuse one either).
- No. A triangle with angles 60^\circ, 50^\circ and 70^\circ has one 60^\circ angle, but no two angles are equal, so by the Base Angles Theorem no two sides are equal either. (If two sides were equal, the angles opposite them would be equal.)
- Isosceles means at least two congruent sides. An equilateral triangle has three, so any two of them qualify. The converse fails because an isosceles triangle may have its third side a different length, for example sides 5, 5 and 8.
- The sides opposite \angle A and \angle B are \overline{BC} and \overline{AC}, not \overline{AB}, which lies between the two angles. The conclusion should be AC = BC.
- The Base Angles Theorem assumes two sides are congruent, and AB = AC is what the proof sets out to show. Quoting it would assume the conclusion.
- A congruence statement is a matching of vertices, and \triangle ABC and \triangle ACB are two different matchings of the same three points: A with A, B with C, C with B. SAS only asks that the matched parts be congruent, and here they are: \overline{AB} with \overline{AC}, \angle A with itself, \overline{AC} with \overline{AB}. Nothing requires the two triangles to be in different places.
- By the Base Angles Theorem, \angle ABC \cong \angle ACB. \angle ABE and \angle ABC form a linear pair, so they are supplementary; likewise \angle ACF and \angle ACB. Supplements of congruent angles are congruent (Proving supplements of congruent angles are congruent), so \angle ABE \cong \angle ACF.
- AD = AE, so by the Base Angles Theorem in \triangle ADE, \angle ADE \cong \angle AED. The parallel lines give corresponding angles \angle ADE \cong \angle B (transversal \overleftrightarrow{AB}) and \angle AED \cong \angle C (transversal \overleftrightarrow{AC}). By the symmetric and transitive properties, \angle B \cong \angle C, and by the converse of the Base Angles Theorem, AB = AC.
- \overrightarrow{AF} bisects \angle CAE, so \angle EAF \cong \angle FAC. Since \overrightarrow{AF} \parallel \overline{BC}, \angle EAF \cong \angle B (corresponding angles, transversal \overleftrightarrow{BE}) and \angle FAC \cong \angle C (alternate interior angles, transversal \overleftrightarrow{AC}). By the symmetric and transitive properties, \angle B \cong \angle C, so AB = AC by the converse of the Base Angles Theorem.
- \angle A is opposite \overline{BC} and \angle B is opposite \overline{AC}, so the converse of the Base Angles Theorem turns \angle A \cong \angle B into \overline{BC} \cong \overline{AC}. With \overline{AB} \cong \overline{AC} (given) and the symmetric and transitive properties, \overline{AB} \cong \overline{AC} \cong \overline{BC}, so the triangle is equilateral. No angle measure was needed.