Exterior Angle Inequality

Department of Applied Mathematics ISU Geometry Spring 2008 1
Lecture 7 – Lecture note by Jerry Yeh from Textbook Section 3.4. – April 8, 2008
Exterior Angle Inequality
Definition 1 Let △ABC be given, and suppose D is a point on ←→ BC such that the betweenness
relation B − C − D holds. Then ∠ACD is called an exterior angle of the given triangle. The
angles at A and B of △ABC are called opposite interior angles of ∠ACD.
A
✁ ✁✁❅ ❅❅
❅❅❅
✁ ✁✁
B
C
D
Note that the third angle of the triangle, ∠ACB forms a linear pair with the exterior angle
∠ACD.
Theorem 2 (The Exterior Angle Inequality) The exterior angle of a triangle has angle measure
greater than that of either opposite interior angle. That is , given △ABC and point D such that
B − C − D, (a) m∠ACD > m∠A and (b) m∠ACD > m∠B.
(a)
✟✯
A
E
❅ M ❅❅❅❅
✁ ✁✁✁✁✁❅ ✟
✟
✟✟✟✟✟✟✟✟✟✟✟
✁ ✁✁✁✁✁
B
C
D
(b)
A
✁ ✁✁✁❅ ❅
❅❅
B C❅ D ❅❅
F
Proof. (a) Locate M, the midpoint of AC. Extend BM to a point E on it such that M is
the midpoint of BE. We then have △AMB ∼ = △CME by SAS. Thus, m∠ACE = m∠MCE =
m∠MAB = m∠A and so, m∠ACD = m∠ACE + m∠ECD > m∠ACE = m∠A.
(b) For the second part, merely locate F so that A − C − F ; then m∠BCF = m∠ACD by the Vertical
Pair Theorem, and from the first part, m∠BCF > m∠B. By substitution, m∠ACD > m∠B.
The proofs of the following Corollaries are left as exercises.
Corollary 3 The sum of the measures of any two angles of a triangle is less than 180
Corollary 4 A triangle can have at most one right or obtuse angle.
Corollary 5 Base angles of an isosceles triangle are acute.
Definition 6 The angle sum for a triangle is the sum of the measures of its three angles.
Theorem 7 (SACCHERI-LEGENDRE Theorem) The angle sum of any triangle cannot exceed
180

Department of Applied Mathematics ISU Geometry Spring 2008 2
Proof. Begin with △ABC and the construction used in the proof of the exterior Angle Inequality :
Find the midpoint M of AC and find E so that B−M −E and BM = ME; connect C and E to form
△BEC. As in the proof of Theorem reftheorem:EXTERIOR, SAS gives us △AMB ∼ = △CME
and hence
m∠ABE = m∠BEC and m∠BAC = m∠ACE
Thus, in the triangle △BCE,
m∠EBC + m∠BCE + m∠BEC = m∠EBC + m∠BCA + m∠ACE + m∠ABE
= m∠EBC + m∠BCA + m∠BAC + m∠ABE
= m∠ABC + m∠BCA + m∠BAC
That is, the angle sum of △ABC is equal to the angle sum of △BCE.
We can then repeat the construction on △BEC : locate the midpoint N of CE and extend BN
its own length to point F such that B −N −F . Continue this process. Here is what happens. First,
the angle sums of all the new triangles we constructed in the process (△BCE, △BCF, △BCG, · · · )
remain constant. Also, the construction allows us to use SAS to show that m∠E = ∠ABE, m∠F =
m∠EBF, m∠G = m∠F BG · · · . We will call these angle measures θ 1 , θ 2 , θ 3 , · · · .
Now, let’s suppose that, contrary to what we are trying to prove, the angle sum of △ABC is
greater than 180. Hence there is some positive constant t such that
m∠A + m∠ABC + m∠BCA = 180 + t
Then all the other triangles we constructed will also have angle sum = 180 + t. Now, observe
that
θ 1 < m∠ABC, θ 1 + θ 2 < m∠ABC, θ 1 + θ 2 + θ 3 < m∠ABC.
In fact, for any n,
θ 1 + θ 2 + θ 3 + · · · + θ n < m∠ABC.
This means ultimately that for at least one n taken large enough, θ n < t. (for if this were not true,
then θ n ≥ t for any n, we would have
m∠ABC > θ 1 + θ 2 + θ 3 + · · · + θ n ≥ nt for any n.
This is impossible.)
So suppose, for convenience, that the triangle for which this occurs is labeled BW C and that
m∠W < t. The angle sum of △BW C = 180 + t, so it follows that
180 + t = m∠W BC + m∠BCW + m∠W < m∠W BC + m∠BCW + t
or, canceling t from both sides,
a contradiction to Corollary 3.
180 < m∠W BC + m∠BCW,