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MTH 250 Graded Assignment 6, part one
Quadrilaterals
Material from Kay, section 3.7
Q1: Various congruence theorems for convex quadrilaterals have been proven: SASAS, SASAA, and
SASSS. There's one more: ASASA. State and prove the ASASA congruence theorem in a manner
similar to the statements (and proofs) of the previous theorems.
If under the correspondence ABCD  XYZW we have
A
X
AB  XY
B
Y
BC  YZ
C
Z
then ABCD  XYZW (i.e., we need to show
D
W , CD  ZW , DA  WX ).
Q2: In the quadrilateral ABCD , AD  BC and A  B .
a) Prove AC  BD
b) And use this result to prove that the diagonals of a
Saccheri Quadrilateral are congruent.
[Kay, Section 3.7, #10, p.192]
Note that in part (a), you prove that the diagonals of a quadrilateral with congruent base angles and
sides are congruent; part (b) immediately extends this to proving that the diagonals of a Saccheri
quadrilateral are congruent. By definition, a Saccheri quadrilateral has congruent base angles and
sides, and so part (a) simply applies to it and you get the result. So, really, there’s nothing to prove
for (b) – just prove part (a)!
Q3: A property of the Saccheri quadrilateral stated on p. 187 is that the line joining the midpoints of
the base and summit is the perpendicular bisector of both the base and the summit. Prove it.
Given: Saccheri quadrilateral ABCD with base AB , E is the midpoint
of AB , and F is the midpoint of CD .
Hints: separate the "bisector" part and the "perpendicular" part. You get that the segments are
bisected immediately- by definition [p 85] anything that passes through the midpoint of a segment
bisects it. State that and move on – you need to get right angles at E and F.

Try working towards meeting the conditions of Theorem 3 on page 98.

To do that, you need to show

FEA  FEB and EFD  EFC .
And to do THAT, try proving that AEFD  BEFC . With the givens and the
theorem about summit angles, you've easily got an SASAS. So prove that first, and then
take what you need from CPCF.
Q4: Another property of the Saccheri quadrilateral stated on page 187 is that if each of the summit
angles is a right angle (i.e. assuming we are in a geometry where such a thing is possible), then the
quadrilateral is a rectangle, and the summit is congruent to the base.
Again, you get part of this immediately – a rectangle is defined as a convex quadrilateral with four
right angles. By definition of a Saccheri quadrilateral, the base angles are right angles, and now, it is
given that the summit angles are as well. That part's done. What you need to prove is that the
summit is congruent to the base.
Hint: Try considering a quadrilateral with the givens as shown, with the diagonals drawn in.
Can you get ABD  CDB ? Note that SSA is unambiguous for a right triangle – the HL or
HA corollaries on pp 176-177 get you there fast. After that, you’ve pretty much got what you
need.
Q5: Problem #8, on p 191 of Kay: In the Lambert Quadrilateral ABCD shown (take givens as
marked, be sure to state them), prove that AD  CD and ray BD bisects ABC .
Hints:


Work first to get the right triangles congruent so you can claim parts.
Be sure to look at the precise definition of an angle bisector on p 93 – you need to formally
assert that ray BD lies between ray BA and ray BC , in addition to what you usually think
of as "definition of bisector" – that the two halves of the angle have equal measures. This is
one of those "looks obvious but is a pain to set up formally" things – you have to look at the
dependencies in the definitions to do it right! In particular:
o
o
o
You can assert D is an interior point of ABC by property of convex quadrilateral.
Since D is an interior point, angle addition holds (look at p 91)
And since angle addition holds, you get the definition of betweenness for rays.
o
That PLUS the fact that you've got (hopefully!)
bisector.
m ABD  m CBD is the precise definition of