Download Math 361 ACTIVITY 8: Following Saccheri —eliminating the obtuse

Math 361
ACTIVITY 8: Following Saccheri —eliminating the obtuse angle hypothesis
Why
In this activity you will complete the first part of Saccheri’s program - showing that [in a neutral
geometry] the summit angles in a Sacceri quadrilateral cannot be obtuse angles.
LEARNING OBJECTIVES
1. Continue to develop skill in working as a team
2. Develop further skill in working with a formal system of geometry
3. Understand more of Saccheris work on non-Euclidean geometry
CITERIA
1. Success in working as a team in generating ideas, correcting errors, and developing understanding
2. Success in completing the exercise
3. Accuracy and completeness (and explanation) of your proofs
RESOURCES
1. Your text
2. The outline notes
3. 40 minutes
PLAN
1. Select roles, if you have not already done so, and decide how you will carry out steps 2 and 3 (5
minutes)
2. Work through the exercises given here - be sure everyone understands all results (30 minutes)
3. Assess the team’s work and roles performances and prepare the Reflector’s and Recorder’s reports
including team grade (5 minutes).
4. Be prepared to discuss your results.
EXERCISE
A Saccheri quadrilateral is a quadrilateral 2ABCD in which C and D are on the same side of
←→
AB AD and BC are perpendicular to AB, and AD ∼
= BC (see Figure 1). Segment AB is the base,
segments AD and BC are the legs and CD is the summit. Saccheri sough to prove Euclid’s Fifth
(parallel) postulate as a theorem in neutral geometry by showing that such a quadrilateral is
a rectangle — a quadrilateral with four right angles. To do this, he intended to show that the
summit angles 6 ADC and 6 BCD are neither obtuse angles nor acute angles.
Figure 1: Saccheri quadrilateral 2ABCD
Your job is to show that the summit angles are not obtuse, through the following program
[Prove the statements marked — mostly these are not one-step proofs. I’ve given some hints]
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1. (Background that you do not have to prove) Since a Saccheri quadrilateral is convex, the SASAS
criterion for convex quadrilaterals shows that 2ABCD ∼
= 2BADC so 6 ADC ∼
= 6 BCD. [The summit
angles of the Saccheri quadrilateral are congruent. Thus both are obtuse or both are acute or both
are right angles].
2. We construct a triangle whose angle sum is related to the summit angles as follows:
(a) We construct the perpendicular bisector of AB, meeting AB at the midpoint M , and locate the
←→
point P on the perpendicular bisector for which P is on the opposite side of AB from C, D and
M P = AD = BC. [We know every segment has a perpendicular bisector and a midpoint. Segment
construction allows us to say that P is determined.]
←→
(b) The segment DP meets AB at a point Z between A and M .
PROVE THIS [based on plane separation, used three times — to get an intersection (Z), to
←→
←→
show the intersection Z is on the M side of AD and (similarly) on the A side of M P (so Z is
←→
←→
between between A and M ) – this requires knowing that all of M P is on the same side of AB,
and vice versa ]
←→
Similarly, [You do not have to write out the proof again] CP meets AB at a point W between M
and B.[See Figure 2]
3. 4DAZ ∼
= 4P M Z [and, similarly, 4CBW ∼
= 4P M W ] so (by the definition of “congruent triangles”)
m6 ADZ = m6 M P Z [and m6 BCW = m6 M P W ]
PROVE THIS (using a recently discussed congruence criterion –it is important that Z is between A
and M and that W is between B and M )
4. The angle sum of 4DP C is equal to m6 ADC + m6 BCD
PROVE THIS [You need to check the conditions for angle addition, using definitions & theorems
concerning interiors of angles]
5. m6 ADC + m6 BCD ≤ 180 PROVE THIS - it’s short, but involves an important fact.
6. The angles 6 ADC and 6 BCD cannot be obtuse angles.
PROVE THIS - it’s pretty trivial
Figure 2: Saccheri quadrilateral 2ABCD
SKILL EXERCISES:(hand in - individually - with this week’s assignments)
Write: text p. 190 (section 3.7) # 4, 10, 20, 21
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