Download MATH 402 Worksheet 8

MATH 402 Worksheet 8
Friday 10/21/16
In the project about Saccheri quadrilaterals, you learned that in a Saccheri quadrilateral as in the figure
below (with the angles at A and B being right), you also have:
• The angles at C and D are acute and equal, and
• The lines AB and CD are ultraparallel to each other, as they share a common perpendicular, which
is 7.5.
the perpendicular
to both segments AB
andTRIANGLES
CD.
LAMBERTbisector
QUADRILATERALS
AND
291
F
D
C
E
A
B
Fig. 7.5
There is another special type of a quadrilateral in hyperbolic geometry: that’s a quadrilateral with 3 right
angles, which Let
is called
a Lambert
E and
F be quadrilateral.
the midpoints of the base and summit of a Saccheri
quadrilateral.
By the
results of the
last section,
we be
know
that
and we get
(1) Show
that in a Lambert
quadrilateral,
the fourth
angle must
acute.
As a ADF
consequence,
⇠
are do
congruent
SAS and
thus AF = BF . Thus, AEF and
that BCF
rectangles
not exist inbyhyperbolic
geometry.
Hint:
First
that youby
canSSS
“double”
a Lambert
quadrilateral
to a
getright
a Saccheri
BEF
areprove
congruent
and the
angle at
E must be
angle.quadrilateral.
A
◦ , using the following
(2) Show
that
the
angle
sum
of
a
hyperbolic
triangle
4ABC
is
strictly
less
that
180
similar argument will show that \DF E and \CF E are also right angles.
strategy:
Thus,
• Assume 4ABC is a right triangle, with right angle at A. Take the midpoint M of BC, and
drop a perpendicular
AC fromjoining
M at D.the
Find
a Lambertofquadrilateral
withsummit
vertices A, B, D.
Theorem
7.9. The to
segment
midpoints
the base and
• Use knowledge about the angles of this Lambert quadrilateral to show the angle sum of 4ABC
of isa less
Saccheri
quadrilateral makes right angles with the base and summit.
than 180◦ .
• Given
general
triangle
necessarily right),
into two
right both
triangles
and use the
If wealook
at the
two (not
quadrilaterals
AEF Dsplit
andit BEF
C, they
share
above.
the that
property
of having
three
right angles.
(3) Show
the angle
sum of any
quadrilateral
is less than 360◦ .
Definition
7.10.between
A Lambert
Quadrilateral
quadrilateral
three
Given 4ABC,
the difference
180◦ and
the angle sumisofa 4ABC
is called having
the defect
of 4ABC.
◦
right angles.
Given a quadrilateral
ABCD, its defect is defined to be the difference between 360 and its angle sum.
The midpoint
construction
just described
gives acorresponding
natural way to
associate
(4) Use defects
to show that
if two triangles
have congruent
angles,
then they are
congruent.
a Lambert quadrilateral with a given Saccheri quadrilateral. We can also
create a Saccheri quadrilateral from a given Lambert quadrilateral.
[Bonus:] Use Saccheri quadrilaterals to show that parallel lines cannot be everywhere equidistant.
Theorem 7.10. Let ABDC be a Lambert quadrilateral with right angles
at A, B, and C. If we extend AB and CD, we can find points E and F
such that AB ⇠
= AE and CD ⇠
= CF (Fig. 7.6). Then EBDF is a Saccheri
quadrilateral.
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