For the next few problems, suppose we are operating under Saccheri's Hypothesis of the Acute Angle (HAA). That is, we suppose that the congruent summit angles of any Saccheri Quadrilateral are acute. The thrust of these problems is to prove (1) that under HAA, triangles have fewer than 180° and (2) that under HAA the Pythagorean relationship between hypotenuse and legs of a right triangle cannot hold. This development is taken from Richard Trudeau's The Non-Euclidean Revolution. Under HAA, prove that if ABCD is a Saccheri Quadrilateral as shown, then AB -CD. A 23

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter6: Circles
Section6.3: Line And Segment Relationships In The Circle
Problem 39E: The center of a circle of radius 2 in. is at a distance of 10 in. from the center of a circle of...
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For the next few problems, suppose we are operating under Saccheri's Hypothesis of the Acute
Angle (HAA). That is, we suppose that the congruent summit angles of any Saccheri
Quadrilateral are acute. The thrust of these problems is to prove (1) that under HAA, triangles
have fewer than 180° and (2) that under HAA the Pythagorean relationship between hypotenuse
and legs of a right triangle cannot hold. This development is taken from Richard Trudeau's The
Non-Euclidean Revolution.
Under HAA, prove that if ABCD is a Saccheri Quadrilateral as shown, then AB - CD.
D,
A
B
%23
Transcribed Image Text:For the next few problems, suppose we are operating under Saccheri's Hypothesis of the Acute Angle (HAA). That is, we suppose that the congruent summit angles of any Saccheri Quadrilateral are acute. The thrust of these problems is to prove (1) that under HAA, triangles have fewer than 180° and (2) that under HAA the Pythagorean relationship between hypotenuse and legs of a right triangle cannot hold. This development is taken from Richard Trudeau's The Non-Euclidean Revolution. Under HAA, prove that if ABCD is a Saccheri Quadrilateral as shown, then AB - CD. D, A B %23
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