Download Material on absolute geometry

MTH 250 Graded Assignment 5 (part 3)
Theorems of absolute geometry (questions)
Material from Kay, sections 3.3, 3.4, 3.5
Q1 (long question, short answer!): Saccheri-Legendre, which states that the sum of the angles
in a triangle in absolute geometry cannot exceed 180, is one of the key theorems of absolute
geometry. Its proof involves building up several results. Just to refresh your memory, the first
result comes out of the proof of the Exterior Angle Inequality. In particular, there’s a specific
construction that takes place, with the result that in the figure below
Next, there’s a key lemma, which states:
Take a triangle ABC and construct a segment BE from B through the midpoint M of AC ,
so that BM  ME . Form a triangle BCE . [This is the same construction used in the exterior
angle inequality.]
Prove that the sum of the angles in the triangle formed by this construction, BCE , is
equal to the sum of the angles of ABC .
That’s the goal of this question – finish the proof of this lemma. Use the setup described in the
lecture, assigning variables:
And take as a given that AMB  CME [you don’t need to re-prove this part, since it’s already
been proven – use it as the jumping off point]. All that remains to be shown is that sum of the
angles in BCE is equal to the sum of the angles in ABC .
Q2: There’s a collection of lemmas about perpendiculars and bisectors related to the symmetry
of an isosceles triangle (they come out of the Isosceles Triangle Theorem, and are used to get to
the Perpendicular Bisector Theorem). They’re familiar results that are straightforward to prove,
and Lemma B and Lemma C are done in the suggested problems. What’s left is Lemma D (p
144), which states
If PA  PB and M is the midpoint of segment AB , then ray
PM bisects APB .
Fill in the proof of Lemma D:
Proof: Since PA  PB (equivalent PA  PB ),
APB is by definition
isosceles, with base AB . By the Isosceles Triangle Theorem,
[Fill in 1]
Since M is the midpoint of segment AB , by definition of midpoint,
[Fill in 2]
Therefore, by the [Fill in 3] , AMP  BMP .
Make a statement about angles APM and BPM (and justify):
[Fill in 4] by [Fill in 5]
Next, since M  AB , M is an interior point of
angle addition holds and [Fill in 6]
APB . Since M is an interior point of APB ,
The above gives us the definition of betweenness for rays: PA  PM  PB .
And finally, PA  PM  PB along with m APM  m BPM is the precise [Fill in 7]
Therefore, ray PM bisects APB .
*Note: Since the structure of this one is already laid out, I'm looking closely at
notation/precision of statements.
Q3: In the figure to the right, it is given that WS  WT , and
RS  ST  TU with R  S  T U . Prove that
RWS  TWU . [Kay, Section 3.3 #18, p. 151]
As always, be sure to write justifications within Kay’s
axiomatic system, and be particular about the fine details.
For example (hey, here’s a hint!), it’s obvious “by looks” that
WST and WSR form a linear pair, but to set that up,
you first need to assert that ST and SR are opposite rays
since R  S  T . That level of picky.
Q4: Use the Exterior Angle Inequality to give upper and
lower bounds for x and y in the diagram. Be sure to
justify your calculations (you can assume betweenness,
opposite rays, and so on as it appears in the diagram
for this one without it being explicitly given).
Q5: In SKL we already know that
m1  m2  180 by a corollary of the Exterior Angle
Inequality. Show that if ray KS bisects WKL and ray
LS bisects WLK , then we can further conclude
that m1  m2  90 . [Kay, Section 3.4 #18, p. 165]
[Relationships as apparent in diagram; 1 is LKS ,
2 is KLS .]
Q6: Given QR  QT and angle measures as indicated in the
diagram, determine which segment in the figure is the shortest, and
justify your answer. [Kay, Section 3.5 #12, p. 172]
Be careful to NOT assume that the sum of angles in the triangle is
exactly 180, because that’s a theorem of Euclidean geometry…and
we aren’t in Euclidean geometry. There are two key theorems of
Section 3.5 at work here – be sure to cite as used.