In elliptic geometry prove the length of the summit of a saccheri quad is less t
ID: 1946849 • Letter: I
Question
In elliptic geometry prove the length of the summit of a saccheri quad is less than the base Given: DA BCExplanation / Answer
Theorem: The summit of a Saccheri Quadrilateral is longer than the base. Proof: Quadrilateral FCBE is constructed (F is the midpoint of DC and E is the midpoint of AB) so that it is a Lambert quadrilateral, then since the side adjacent to the acute angle of a Lambert quadrilateral is longer than the opposite side we know that FC > EB. Similarly, FDAE is a Lambert quadrilateral so that FD > EA. It follows that FC + FD > EA + EB which is the same as saying DC > AB. QED If needed. Theorem: In a Lambert quadrilateral where angle A is the acute angle AD is longer than CB (the side adjacent to the acute angle is longer than the opposite side). Proof by contradiction: Case 1: Assume AB is congruent to DC, then we have a Saccheri quadrilateral in ABCD, but the measure of angle D is 90 degrees, contradicting that it must be acute. Case 2: CD is longer than AB. Then construct point E so that C-E-D (E is between C and D) and so that ABCE is a Saccheri Quadrilateral where angles C and B are the right angles and EC = AB. Then the measure of angle AEC is less than 90 degrees, but it is also an exterior angle of triangle EDA so that measure of angle D is equal to 90 degrees which is less than measure of angle AEC making AEC obtuse, a contradiction (earlier it was shown that AEC must be acute because it is a summit angle).