Geometry · G.CO.C.10

Proving the base angles theorem

The Base Angles Theorem needs two triangles to compare, and an isosceles triangle is only one. Which line cuts it into two halves that can be proved congruent, and why may that line be drawn?

You need

Isosceles and equilateral trianglesGeometry Proving parts congruent after the trianglesGeometry Auxiliary linesGeometry

Where you are 3 in Proving triangles congruent

Geometry11

Geometric proof18

Proving triangles congruent3

Proving parts congruent after the triangles

Proving the base angles theorem

Planning a congruence proof

When you reach for it

The Base Angles Theorem says that if two sides of a triangle are congruent, the angles opposite them are congruent. That lesson proves it by Pappus’s method, matching the triangle with itself, and draws no line. This lesson gives the two proofs most textbooks use. Each adds one auxiliary line from the vertex to the base, cutting the triangle into two halves, and proves the halves congruent.

Reach for them when a proof needs more than the base angles: the congruent halves also show that the added line halves the base, halves the vertex angle, and meets the base at right angles. It is proving parts congruent after the triangles, with the triangles made first.

The procedure

Given. \overline{AB} \cong \overline{AC} in \triangle ABC.

Prove. \angle B \cong \angle C.

Figure 1The Given. The base angles carry no arcs, since their congruence is the Prove.

Choosing the line. Three segments from A to \overline{BC} come to mind: the bisector of \angle A, the segment to the midpoint of \overline{BC}, and the perpendicular from A. Each exists and is unique (Auxiliary lines lists why). Any segment \overline{AD} from A to the base makes \triangle ABD and \triangle ACD with two pairs of matching sides already: \overline{AB} \cong \overline{AC} from the Given, and \overline{AD} \cong \overline{AD} by the reflexive property. The line’s definition has to supply a third pair. Once the halves are congruent, the base angles follow, because corresponding parts of congruent triangles are congruent.

The bisector supplies \angle BAD \cong \angle CAD, the angle between the two known sides. That is SAS.
The segment to the midpoint supplies \overline{BD} \cong \overline{CD}, the third sides. That is SSS.
The perpendicular supplies two right angles at D. Each sits opposite a leg, not between the known sides, so the three pairs are side, side and a non-included angle, and SSA is not a criterion. Example 3 takes this up.

Proof 1, with the bisector (SAS).

Figure 2Proof 1 adds the bisector of \angle A. The arcs at A are allowed, because the line was made that way.

Statements	Reasons
1. \overline{AB} \cong \overline{AC}	given
2. The bisector of \angle BAC meets \overline{BC} at a point D between B and C	an angle has exactly one bisector; Crossbar Theorem, a position fact drawn in Figure 2
3. \angle BAD \cong \angle CAD	definition of angle bisector: line 2
4. \overline{AD} \cong \overline{AD}	reflexive property of congruence
5. \triangle ABD \cong \triangle ACD	SAS postulate: lines 1, 3 and 4
6. \angle B \cong \angle C	corresponding parts of congruent triangles: line 5

The bisector runs inside \angle BAC, and the Crossbar Theorem, named in proving the exterior angle theorem, says a ray from an angle’s vertex through its inside crosses every segment joining the angle’s two sides. As there, it is read from the figure (what a diagram may assume); where the bisector crosses is a measure fact, and line 2 claims nothing about it. Line 5 matches A \leftrightarrow A, B \leftrightarrow C, D \leftrightarrow D, with \angle BAD between \overline{AB} and \overline{AD}, as SAS needs. In line 6, \angle ABD is the whole triangle’s \angle B, because D lies on \overline{BC}, and likewise at C.

Proof 2, with the midpoint (SSS).

Figure 3Proof 2 adds the segment to the midpoint of the base. The ticks on \overline{BM} and \overline{MC} are allowed for the same reason.

Statements	Reasons
1. \overline{AB} \cong \overline{AC}	given
2. M is the midpoint of \overline{BC}, joined to A by \overline{AM}	a segment has exactly one midpoint, and two points determine a segment
3. \overline{BM} \cong \overline{CM}	definition of midpoint: line 2
4. \overline{AM} \cong \overline{AM}	reflexive property of congruence
5. \triangle ABM \cong \triangle ACM	SSS postulate: lines 1, 3 and 4
6. \angle B \cong \angle C	corresponding parts of congruent triangles: line 5

Neither proof cites the Base Angles Theorem, or anything proved from it.

What else the halves give. Once \triangle ABD \cong \triangle ACD, the other pairs of parts are congruent too: \overline{BD} \cong \overline{CD}, so the bisector lands on the midpoint of the base, and \angle ADB \cong \angle ADC, a linear pair of congruent angles, so each is a right angle. Proof 2 gives the same from the other side. Put together:

In an isosceles triangle, the bisector of the vertex angle, the segment from the vertex to the midpoint of the base, and the perpendicular from the vertex to the base are one segment.

Examples 1 and 2 write the steps out, and Exercise 18 explains why they are one segment.

Worked examples

Example 1

The bisector halves the base

Given. \overline{AB} \cong \overline{AC}; \overline{AD} bisects \angle BAC and meets \overline{BC} at D, as in Figure 2.

Prove. D is the midpoint of \overline{BC}, and \overline{AD} \perp \overline{BC}. Then, with BD = 2x + 3 and DC = 5x - 9, find x and BC.

Solution

Plan. Prove the halves congruent exactly as in Proof 1, then collect two more pairs of parts.

Statements	Reasons
1. \overline{AB} \cong \overline{AC}	given
2. \overline{AD} bisects \angle BAC	given
3. \angle BAD \cong \angle CAD	definition of angle bisector: line 2
4. \overline{AD} \cong \overline{AD}	reflexive property of congruence
5. \triangle ABD \cong \triangle ACD	SAS postulate: lines 1, 3 and 4
6. \overline{BD} \cong \overline{CD}	corresponding parts of congruent triangles: line 5
7. D is the midpoint of \overline{BC}	definition of midpoint: line 6, D is between B and C
8. \angle ADB \cong \angle ADC	corresponding parts of congruent triangles: line 5
9. \angle ADB and \angle ADC form a linear pair	definition of linear pair: D is between B and C
10. \overline{AD} \perp \overline{BC}	equal neighbors: lines 8 and 9

Line 10 cites the result that the lesson on perpendicular lines calls “equal neighbors”: two angles of a linear pair measure 180^\circ together, so if they are congruent each is 90^\circ, and the lines are perpendicular.

For the numbers, line 6 gives BD = DC, so

\begin{aligned} 2x + 3 &= 5x - 9 \\ 12 &= 3x \\ x &= 4 \end{aligned}

Then BD = 2(4) + 3 = 11 and DC = 5(4) - 9 = 11, so BC = 22.

Example 2

The median halves the vertex angle

Given. \overline{AB} \cong \overline{AC}; M is the midpoint of \overline{BC}, as in Figure 3.

Prove. \overline{AM} bisects \angle BAC. Then, with m\angle BAM = (3y + 4)^\circ and m\angle CAM = (5y - 12)^\circ, find y and all three angles of \triangle ABC.

Solution

Plan. Proof 2’s triangles again, this time read at A.

Statements	Reasons
1. \overline{AB} \cong \overline{AC}	given
2. M is the midpoint of \overline{BC}	given
3. \overline{BM} \cong \overline{CM}	definition of midpoint: line 2
4. \overline{AM} \cong \overline{AM}	reflexive property of congruence
5. \triangle ABM \cong \triangle ACM	SSS postulate: lines 1, 3 and 4
6. \angle BAM \cong \angle CAM	corresponding parts of congruent triangles: line 5
7. \overline{AM} bisects \angle BAC	definition of angle bisector: line 6, \overline{AM} is inside \angle BAC

Line 6 and the definition of congruent angles give 3y + 4 = 5y - 12, so 16 = 2y and y = 8. Each half is 3(8) + 4 = 28 degrees, so m\angle BAC = 56^\circ, and the base angles share what is left:

m\angle B = m\angle C = \frac{180^\circ - 56^\circ}{2} = 62^\circ

Check. 5(8) - 12 = 28, and 56 + 62 + 62 = 180.

Example 3

The perpendicular, and a proof that fails

Given. \overline{AB} \cong \overline{AC}; \overline{AH} \perp \overline{BC} at H, as in Figure 4.

Someone offers this proof of \angle B \cong \angle C: “\angle AHB \cong \angle AHC since both are right; \overline{AH} \cong \overline{AH}; \overline{AB} \cong \overline{AC}; so \triangle ABH \cong \triangle ACH by SAS, and \angle B \cong \angle C.” What is wrong? Then, with the theorem proved another way, show that H is the midpoint of \overline{BC}, and find AH when AB = AC = 25 and BC = 14.

Figure 4The perpendicular from A. It exists, but what it supplies does not fit SAS or SSS.

Solution

SAS needs the angle between the two sides. In \triangle ABH the angle between \overline{AH} and \overline{AB} is \angle BAH; the right angle at H sits opposite \overline{AB}. So the parts are side, side and a non-included angle, and SSA proves nothing. The conclusion is true; the reason is not.

The criterion that fits is HL. This map proves HL from the Pythagorean theorem and SSS, without this theorem, so the altitude does give a third proof, written out as Example 4 of that lesson. (Courses that prove HL from the Base Angles Theorem cannot use it here: that would be circular.)

Once the theorem is proved, the perpendicular is fine to use. In \triangle ABH and \triangle ACH: \angle B \cong \angle C by the Base Angles Theorem, \angle AHB \cong \angle AHC by the Right Angle Congruence Theorem, and \overline{AB} \cong \overline{AC}. The side \overline{AB} does not join B to H, so it is not between those two angles, and the AAS theorem gives \triangle ABH \cong \triangle ACH. Then \overline{BH} \cong \overline{CH} as corresponding parts, and H is the midpoint. (H is between B and C because the base angles, being congruent, are both acute.)

So BH = 14 \div 2 = 7, and the Pythagorean theorem in the right triangle ABH gives

AH = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24

Example 4

A kite, split along its diagonal

Given. ABCD is a quadrilateral with \overline{AB} \cong \overline{AD} and \overline{CB} \cong \overline{CD}, as in Figure 5.

Prove. \angle ABC \cong \angle ADC. Then, with m\angle BAD = 40^\circ and m\angle BCD = 100^\circ, find m\angle ABC.

Figure 5A kite. The diagonal \overline{AC} cuts it into two triangles that share a side.

Solution

Plan. No triangle in the figure contains both angles, so make two. The diagonal \overline{AC} cuts the quadrilateral into \triangle ABC and \triangle ADC, which share it.

Statements	Reasons
1. \overline{AB} \cong \overline{AD}	given
2. \overline{CB} \cong \overline{CD}	given
3. \overline{AC} is drawn	two points determine a segment
4. \overline{AC} \cong \overline{AC}	reflexive property of congruence
5. \triangle ABC \cong \triangle ADC	SSS postulate: lines 1, 2 and 4
6. \angle ABC \cong \angle ADC	corresponding parts of congruent triangles: line 5

As in Proof 2, the added segment is a shared side, and line 5 matches A \leftrightarrow A, B \leftrightarrow D, C \leftrightarrow C.

The numbers come from the other diagonal. \overline{BD} makes two isosceles triangles, and the Base Angles Theorem, now proved, applies to each. In \triangle ABD the base angles are (180^\circ - 40^\circ) \div 2 = 70^\circ; in \triangle CBD they are (180^\circ - 100^\circ) \div 2 = 40^\circ. \overline{BD} lies inside \angle ABC, so

m\angle ABC = 70^\circ + 40^\circ = 110^\circ

Check. 40 + 100 + 110 + 110 = 360, the angle sums of the two triangles that \overline{BD} makes.

Example 5

A point on the bisector

Given. \overline{AB} \cong \overline{AC}; \overline{AD} bisects \angle BAC and meets \overline{BC} at D; P lies on \overline{AD} between A and D, as in Figure 6.

Prove. \angle PBC \cong \angle PCB. Then, with m\angle ABC = 70^\circ and m\angle ABP = 20^\circ, find m\angle PBC and m\angle BPC.

Figure 6P is any point of the bisector inside the triangle.

Solution

Plan. Split along \overline{AP} instead of \overline{AD}. That gives \overline{PB} \cong \overline{PC}, so \triangle PBC is isosceles, and the theorem finishes.

Statements	Reasons
1. \overline{AB} \cong \overline{AC}	given
2. \overline{AD} bisects \angle BAC	given
3. \angle BAP \cong \angle CAP	definition of angle bisector: line 2, P is on \overline{AD}
4. \overline{AP} \cong \overline{AP}	reflexive property of congruence
5. \triangle ABP \cong \triangle ACP	SAS postulate: lines 1, 3 and 4
6. \overline{PB} \cong \overline{PC}	corresponding parts of congruent triangles: line 5
7. \angle PBC \cong \angle PCB	Base Angles Theorem: line 6, in \triangle PBC

Line 3 holds because \angle BAP is \angle BAD: P lies on \overrightarrow{AD}. Line 7 may cite the theorem, since this proves something else. Lines 1 to 6 work wherever P sits on the ray, so every point of the bisector is equally far from B and C, a first look at the perpendicular bisector theorem.

For the numbers, \overline{BP} lies inside \angle ABC, so m\angle PBC = 70^\circ - 20^\circ = 50^\circ. By line 7, m\angle PCB = 50^\circ too, and the angle sum of \triangle PBC gives

m\angle BPC = 180^\circ - 50^\circ - 50^\circ = 80^\circ

Boundaries

The added line gets one condition besides passing through A. A bisector, or a midpoint, or a perpendicular, never two at once: “let \overline{AD} bisect \angle A and meet \overline{BC} at its midpoint” assumes what the proof will show. That they turn out to be one line is a result, true only because two sides are congruent.

The perpendicular needs more tools. With SSS, SAS, ASA and AAS it gives only side, side and a non-included angle (Example 3).

Nothing that rests on the theorem may prove it: not equilateral and equiangular, and not HL in a course that proves HL from it. The converse is no help either, since its hypothesis is this Prove.

Only when two sides are congruent. If AB \ne AC, the bisector of \angle A and the segment to the midpoint of \overline{BC} are different lines, and neither makes congruent halves.

Folding is the same idea. The fold in Isosceles and equilateral triangles is a reflection; these proofs use SAS or SSS instead, which can themselves be derived from rigid motions.

Where this goes next

Run Proof 1 backwards, with AAS in place of SAS, and it proves the converse (Exercise 17). In an equilateral triangle the result holds from every vertex, so each angle bisector is also a median and an altitude.

The line found here is the perpendicular bisector of the base, which leads to the perpendicular bisector theorem. The same split gives the properties of a kite, as in Example 4, and of an isosceles trapezoid. In a circle, two radii and a chord form an isosceles triangle, and Example 3 becomes the perpendicular from the center bisects a chord. Euclid’s own proof, comparing two pairs of triangles, is a two-stage proof.

Exercises

Reading the proofs

In Proof 1, which line is the only one that uses the fact that \overline{AD} bisects \angle A? What would go wrong if D were simply “a point of \overline{BC}”?
Write out the correspondence in line 5 of Proof 2, and list the three pairs of sides it matches.
Explain why ASA could not be used with either added line to prove the theorem.

Finding measures

In Figure 2, BD = 3x - 2 and DC = x + 8. Find x and BC.
In Figure 2, m\angle BAD = 34^\circ. Find m\angle B and m\angle ADB.
In Figure 3, m\angle BAM = (4k - 6)^\circ and m\angle MAC = (2k + 14)^\circ. Find k and m\angle C.
In Figure 3, AB = AC = 29 and BC = 40. Find AM.
In Figure 2, m\angle ADC = (5y)^\circ. Find y.
In Figure 4, m\angle B = 38^\circ. Find m\angle BAH and m\angle BAC.
In Figure 5, m\angle BAD = 64^\circ and m\angle BCD = 116^\circ. Find m\angle ABC.
In Figure 6, m\angle ABC = 66^\circ and m\angle PBC = 20^\circ. Find m\angle ACP and m\angle BPC.

Short proofs

Write each as a two-column proof.

Given. \overline{AB} \cong \overline{AC}; M is the midpoint of \overline{BC}. Prove. \overline{AM} \perp \overline{BC}.
Given. In Figure 5, \overline{AB} \cong \overline{AD} and \overline{CB} \cong \overline{CD}. Prove. \angle ABC \cong \angle ADC without the SSS postulate: draw the other diagonal, \overline{BD}, and use the Base Angles Theorem twice.
Given. In Figure 4, \overline{AB} \cong \overline{AC} and \overline{AH} \perp \overline{BC}. Prove. \overline{AH} bisects \angle BAC.

Reasoning

A proof begins: “Let M be the midpoint of \overline{BC}, and let \overline{AM} bisect \angle BAC.” What is wrong, and how should it begin?
In Figure 4 the right angles at H are congruent and \overline{AH} is shared. Name a criterion that would finish a proof of the theorem, and say why this lesson does not use it.
Use the bisector of \angle A and the AAS theorem to prove the converse: if \angle B \cong \angle C in \triangle ABC, then \overline{AB} \cong \overline{AC}. Why would the segment to the midpoint of \overline{BC} not work here?
Explain why, in an isosceles triangle, the bisector of the vertex angle, the segment from the vertex to the midpoint of the base, and the perpendicular from the vertex to the base are one segment.

Answers

Line 3. Without it only two pairs of sides remain, which fits no criterion; for most points D the halves are not even congruent.
A \leftrightarrow A, B \leftrightarrow C, M \leftrightarrow M. The sides are \overline{AB} with \overline{AC}, \overline{BM} with \overline{CM}, and \overline{AM} with itself.
ASA needs two pairs of angles. With the bisector one pair is known, at A; the other would be \angle B \cong \angle C, the Prove, or \angle ADB \cong \angle ADC, known only afterwards. With the midpoint no pair of angles is known.
x = 5 and BC = 26. \overline{BD} \cong \overline{CD} (Example 1), so 3x - 2 = x + 8 and 2x = 10. Each half is 13.
m\angle B = 56^\circ and m\angle ADB = 90^\circ. m\angle BAC = 2(34^\circ) = 68^\circ, so m\angle B = (180^\circ - 68^\circ) \div 2; \overline{AD} \perp \overline{BC} by Example 1. Check in \triangle ABD: 34 + 56 + 90 = 180.
k = 10 and m\angle C = 56^\circ. \overline{AM} bisects \angle BAC (Example 2), so 4k - 6 = 2k + 14 and 2k = 20. Each half is 34^\circ, m\angle BAC = 68^\circ, and m\angle C = (180^\circ - 68^\circ) \div 2.
AM = 21. Proof 2 gives \angle AMB \cong \angle AMC, a linear pair, so both are right angles. With BM = 20, \displaystyle AM = \sqrt{29^2 - 20^2} = \sqrt{441}.
y = 18. \overline{AD} \perp \overline{BC} (Example 1), so \angle ADC is a right angle and 5y = 90.
m\angle BAH = 52^\circ and m\angle BAC = 104^\circ. In the right triangle ABH, m\angle BAH = 180^\circ - 90^\circ - 38^\circ. m\angle C = 38^\circ too (Base Angles Theorem), so m\angle CAH = 52^\circ the same way, and m\angle BAC = 2(52^\circ). Check: 180 - 2(38) = 104.
m\angle ABC = 90^\circ. With diagonal \overline{BD}, m\angle ABD = (180^\circ - 64^\circ) \div 2 = 58^\circ and m\angle CBD = (180^\circ - 116^\circ) \div 2 = 32^\circ, and 58 + 32 = 90. Check: 64 + 116 + 90 + 90 = 360.
m\angle ACP = 46^\circ and m\angle BPC = 140^\circ. m\angle PCB = 20^\circ (Example 5) and m\angle ACB = 66^\circ, so m\angle ACP = 66^\circ - 20^\circ; then m\angle BPC = 180^\circ - 20^\circ - 20^\circ.
Lines (1) to (5) are those of Example 2, with (2) “M is the midpoint of \overline{BC} (given)” and (5) \triangle ABM \cong \triangle ACM (SSS postulate: lines 1, 3 and 4). (6): \angle AMB \cong \angle AMC (corresponding parts of congruent triangles: line 5). (7): \angle AMB and \angle AMC form a linear pair (definition of linear pair: M is between B and C, line 2). (8): \overline{AM} \perp \overline{BC} (equal neighbors: lines 6 and 7).
Line (1): \overline{AB} \cong \overline{AD} (given). (2): \overline{CB} \cong \overline{CD} (given). (3): \overline{BD} is drawn (two points determine a segment). (4): \angle ABD \cong \angle ADB (Base Angles Theorem: line 1, in \triangle ABD). (5): \angle CBD \cong \angle CDB (Base Angles Theorem: line 2, in \triangle CBD). (6): m\angle ABD = m\angle ADB (definition of congruent angles: line 4). (7): m\angle CBD = m\angle CDB (definition of congruent angles: line 5). (8): m\angle ABD + m\angle CBD = m\angle ABC (angle addition postulate: \overrightarrow{BD} is inside \angle ABC). (9): m\angle ADB + m\angle CDB = m\angle ADC (angle addition postulate: \overrightarrow{DB} is inside \angle ADC). (10): m\angle ADB + m\angle CBD = m\angle ABC (substitution: line 6 into line 8). (11): m\angle ADB + m\angle CDB = m\angle ABC (substitution: line 7 into line 10). (12): m\angle ADC = m\angle ABC (substitution: line 9 into line 11). (13): \angle ADC \cong \angle ABC (definition of congruent angles: line 12). (14): \angle ABC \cong \angle ADC (symmetric property of congruence: line 13). This is Example 4’s Prove again, reached through the two isosceles triangles that \overline{BD} makes, the route Example 4 used for its numbers.
Line (1): \overline{AB} \cong \overline{AC} (given). (2): \angle B \cong \angle C (Base Angles Theorem: line 1). (3): \overline{AH} \perp \overline{BC} (given). (4): \angle AHB and \angle AHC are right angles (four right angles: line 3). (5): \angle AHB \cong \angle AHC (Right Angle Congruence Theorem: line 4). (6): \triangle ABH \cong \triangle ACH (AAS theorem: lines 2, 5 and 1). (7): \angle BAH \cong \angle CAH (corresponding parts of congruent triangles: line 6). (8): \overline{AH} bisects \angle BAC (definition of angle bisector: line 7, \overline{AH} is inside \angle BAC).
It gives \overline{AM} a third condition. “Through A” and “through M” already fix the segment, and “bisects \angle BAC” is what Example 2 proves, so assuming it is circular. It should begin: “Let M be the midpoint of \overline{BC}, and draw \overline{AM}.”
HL, the hypotenuse leg theorem. On this map HL is proved from the Pythagorean theorem and SSS, so it would not be circular, but it is a separate lesson and not one this lesson builds on. In courses that prove HL from the Base Angles Theorem, citing it here would be circular.
Let \overline{AD} bisect \angle BAC, meeting \overline{BC} at D (Crossbar Theorem, as in Proof 1). Then \angle BAD \cong \angle CAD (definition of angle bisector), \angle B \cong \angle C (given) and \overline{AD} \cong \overline{AD} (reflexive property). \overline{AD} does not join A to B, so it is not between those two angles, and the AAS theorem gives \triangle ABD \cong \triangle ACD; then \overline{AB} \cong \overline{AC} as corresponding parts. With the midpoint, the known parts are \overline{BM}, \overline{AM} and \angle B, which is not between those sides: side, side, non-included angle, not a criterion.
By Example 1 the bisector \overline{AD} meets \overline{BC} at its midpoint, and a segment has one midpoint, so \overline{AD} is the segment from A to the midpoint. By Example 1 again, \overline{AD} \perp \overline{BC}, and from a point off a line there is exactly one perpendicular to it, so \overline{AD} is also the perpendicular from A. All three descriptions name the one segment \overline{AD}.