Download Theorem 1. (Exterior Angle Inequality) The measure of an exterior

Theorem 1. (Exterior Angle Inequality) The measure of an exterior angle of a triangle is greater than the
mesaure of either opposite interior angle.
−−→
Proof: Given 4ABC, extend side BC to ray BC and choose a point D on this ray so that C is between B
and D. I claim that m∠ACD > m∠A and m∠ACD > m∠B. Let M be the midpoint of AC and extend the
median BM so that M is the midpoint of BE.
A
E
M
B
D
C
Figure 1.
Then ∠AM B and ∠CM E are congruent vertical angles and 4AM B ∼
= 4CM E by SAS. Consequently,
←→
m∠M CE = m∠M AB = m∠A by CPCTC. Now E lies in the half-plane of A and CD since A and E are
←→
←→
on the same side of CD. Also, E lies in the half-plane of D and AC since D and E are on the same side
←→
of AC. Therefore E lies in the intersection of these two half-planes, which is the interior of ∠ACD. Finally,
m∠ACD = m∠ACE + m∠ECD > m∠ACE = m∠A. The proof that m∠ACD > m∠B is similar and left
to the reader. ¥
Theorem 2. If two lines are cut by a transversal and a pair of alternate interior angles are congruent, the
lines are parallel.
Proof: We prove the contrapositive. Assume that lines l and m intersect at the point R, and suppose that
a transversal t cuts line l at the point A and cuts line m at the point B. Let ∠1 and ∠2 be a pair of alternate
interior angles. Then either ∠1 is an exterior angle of 4ABR and ∠2 is an interior angle opposite to it, or
vise versa.
l
A
1
m
R
2
B
t
Figure 2.
In either case m∠1 6= m∠2 by the Exterior Angle Inequality (Theorem 1). ¥
Note that the converse of Theorem 2 holds in Euclidean geometry but fails in hyperbolic geometry.
Theorem 3. If B − C − D and 4ACD is isoceles with m∠A = m∠D, then m∠ACB ≥ 2m∠D.
Proof: We assume the Euclidean Parallel Postulate and leave the proof in the absolute case to the reader.
Refer to the diagram in Figure 3 and note that m∠ACB, m∠A, and m∠D are independent of the choice of
B.
Figure 3.
So choose B so that C is the midpoint of BD. Then A, B and D lie on the same circle centered at C with
∠ACB and ∠ADB subtending chord AB. Therefore m∠ACB = 2m∠D. ¥
1
−−→
Theorem 4 (The Crossbar Theorem). If point D lies in the interior of ∠BAC, then AD cuts BC at some
point E in the interior of ∠BAC.
Proof: See College Geometry: A Discovery Approach by David C. Kay.
Theorem 5. If the sum of the interior angles of 4ABD is less than 180◦ and C is a point on side BD such
that AC = CD, then the angle sums of 4ABC and 4ACD are less than 180◦ .
Proof: By Lemma 2, the angle sum of 4ABC ≤ 180◦ and the angle sum of 4ACD ≤ 180◦ . If both of
these inequalities were equalities we would have m∠1 + m∠2 + m∠B + m∠D + m∠3 + m∠4 = 360, in which
case the angle sum of 4ABD = m∠1 + m∠2 + m∠B + m∠D = 360 − m∠3 − m∠4 = 180◦ , contradicting
our hypothesis. So at least one of these inequalities is strict. Suppose the angle sum of 4ACD = 180◦ .
Then 2m∠D + m∠4 = 180◦ = m∠3 + m∠4 so that m∠3 = 2m∠D. But ∠1 ≤ 12 m∠4 and ∠B ≤ 12 m∠4
by Theorem 3, and the angle sum of 4ABD = m∠1 + m∠2 + m∠B + m∠D = m∠1 + 2m∠D + m∠B =
m∠1 + m∠B + m∠3 ≤ 12 m∠4 + 12 m∠4 + m∠3 = 180◦ , contradicting our hypothesis. Therefore the angle
sum of 4ACD < 180◦ . Proof that the angle sum of 4ABC < 180◦ is similar and left to the reader. ¥
Definition 1. The defect of 4ABC is 180◦ − m∠A − m∠B − m∠C.
Theorem 6. Every hyperbolic right triangle has positive defect.
Proof: Let 4ABC be a right triangle with right angle at B. Construct the line m through A perpendicular
←→
←→
←→
to AB. Choose a point P on m on the same side of AB as C. Then m is parallel to BC by Theorem 2. By
←→
the Hyperbolic Parallel Postulate, there is another line n through A parallel to BC. Choose a point Q on n
←→
−→
on the same side of AB as C. We may assume that AQ lies in the interior of ∠BAP (if not, consider right
triangle 4ABC 0 ∼
= 4ABC with B the midpoint of CC 0 ). Let t = m∠P AQ.
Figure 4.
−−→
−−→
I claim there is a point W on BC such that m∠AW B < t. Locate points W1 = C, W2 , W3 , . . . on BC such
that B −W1 −W2 and W1 W2 = AW1 , B −W2 −W3 and W2 W3 = AW2 , and so on. Let xn = m∠AWn B; then
1
1
1
2
n
by Theorem
2 ≥ 2 x3 ≥ · · · ≥ 2 xn+1 ≥ · · · so that xn+1 ≤ 2 xn ≤ 22 xn−1 ≤ · · · ≤ 2n x1 . Now if
¡ x1 ¢3, x1 ≥ 2x
1
n > log2 t , then 2n x1 < t. So choose such an integer n and set W = Wn+1 . Then m∠AW B = xn+1 ≤
−−
→ −−→ −→
1
2n x1 < t as claimed. Note that the order of the rays through A must be AB − AW − AQ, for otherwise
−
−→ −→ −−→
−→
AB − AQ − AW , in which case AQ would cut BW by the Crossbar Theorem, contradicting our choice of n.
−
−→ −→ −→
−−
→ −−→ −→ −→
Furthermore, AB − AQ − AP implies AB − AW − AQ − AP and we can now estimate the defect. The sum of
◦
the interior angles of 4ABW = 90 + m∠AW B + m∠BAW < 90◦ + t + m∠BAW < 90◦ + m∠BAP = 180◦
and the conclusion follows from Theorem 5. ¥
Corollary 1. Every hyperbolic triangle has positive defect.
Proof: Label 4ABC so that AC is the longest side. Let D be the foot of the perpendicular from B to AC.
Then 4ABD and 4BCD are right triangles with positive defect. Hence m∠A + m∠ABD + 90◦ < 180◦ and
m∠C + m∠CBD + 90◦ < 180◦ . By adding inequalities we obtain m∠A + m∠B + m∠C < 180◦ . ¥
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