Desargues' Theorem and SSASSDate: 12/15/98 at 18:47:30 From: Brian Walsh Subject: Desargues + congruence postulate Actually I have two questions: 1) Using constructions, determine whether SSASS (Side, Side, Angle, Side, Side) is a congruence postulate for Quadrilaterals. 2) With a ruler, draw 3 rays, OP, OQ, OR. Let A be any point on OP. Let B be any point on OQ. Let C be any point on OR. Draw 6 lines: AB, BC, AC, PQ, PR, and QR. Verify the incredible discovery of Gerard Desargues, in the early 1600's, that either: (a) AB is parallel to PR or BC is parallel to QR or AC is parallel to PQ. Or (b) the 3 points of intersection are collinear. (This result is known as the Desargues Theorem.) Date: 12/16/98 at 03:29:47 From: Doctor Floor Subject: Re: Desargues + congruence postulate Hi Brian, Thanks for your question. For your first question consider the following two figures: * * * * * * * * * * A * A * * * * * * * * * * * Although angle A has the same measure in both of these quadrilaterals, you can see that they have equal SSASS, but are not congruent. I think you will be able to show this using constructions! For Desargues Theorem consider PQR.O as (the projection of) a pyramid. In this pyramid PQR and ABC form two planes. If the two planes are parallel, then you have all three parallelisms of (a). If the two planes intersect, then the intersection line can be parallel to one of the sides of PQR. Then you get one of the parallelisms of (a). In other cases AB and PQ meet at the line of intersection, and so do BC and QR, and AC and PR. And we have case (b). If you have a math question again, please send it to Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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