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Suggested problems - solutions
The Saccheri quadrilateral
Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, 2001. In particular, see section 3.7, pp 190-193.
The problems are all from section 3.7.
Problems: 11, 21
#11:
Prove that if the summit and the base of a Saccheri quadrilateral have equal lengths,
then the quadrilateral is a rectangle.
Let ABCD be a Saccheri quadrilateral with base AB. By definition, AD ∼
= BC, and ∠A and
∠B are right angles.
Also, let it be given that CD = AB (the length of the summit is equal to the length of the base).
Consider the triangles ABD and CDB formed by the diagonal BD.
Since BD ∼
= DB (reflexive), by the SSS theorem, we have that ABD ∼
= CDB. Therefore,
we have corresponding angles ∠A ∼
∠C.
Since
∠A
is
a
right
angle,
∠C
is
a right angle as well.
=
Similiarly, consider the triangles BAC and DCA formed by the diagonal AC.
Since AC ∼
= AC (reflexive), by the SSS theorem, we have that BAC ∼
= DCA. Therefore,
we have corresponding angles ∠B ∼
= ∠D. Since ∠B is a right angle, ∠D is a right angle as well.
The quadrilateral has four right angles, and is thus a rectangle.
#21:
Prove that the base of a Saccheri quadrilateral in absolute geometry has length less
than or equal to that of the summit.
Consider a Saccheri quadrilateral with base AB and diagonal AC as shown:
Since (by property of the Saccheri quadrilateral) C is an interior point of ∠DAB and angle
addition holds, we have
m∠1 + m∠2 = 90
By the Saccheri-Legendre theorem, the sum of the measures of the angles in ABC cannot
exceed 180, so
m∠2 + m∠3 + 90 ≤ 180
m∠2 + m∠3 ≤ 90
Substitution gives
m∠2 + m∠3 ≤ m∠1 + m∠2
so
m∠3 ≤ m∠1
Since AD = CB and AC = CA, we have the conditions set up for the Hinge theorem: so since
m∠3 ≤ m∠1, it must be that AB ≤ CD (the sides opposite those angles).
(base) AB < CD
(summit)