Trapezoid MedianDate: 14 Aug 1995 16:13:12 -0400 From: Damien Mascord Subject: Proof for a geo q Question: PQRS is a trapezium with PQ||SR. If A and B are mid-points of SP and RQ respectively prove that a) AB is parallel to PQ and SR, b) Ab=1/2(PQ + SR) P/----->---\Q / \ A/-------------\B / \ S/---------->------\R Date: 15 Aug 1995 17:22:44 -0400 From: Dr. Ken Subject: Re: Proof for a geo q Hello there! To do this problem, you kind of have to work backwards: instead of saying that A and B are midpoints of PS and RQ, say that A is the midpoint of PS and that line L is the line through A which is parallel to P and S. Since we're in Euclidean Geometry, there's only one such parallel line. Let C be the point at which this line intersects QR (which it does, again because we're in Euclidean Geometry and not something else funky). P/----->---\Q / \ A/-------------\C / \ S/---------->------\R Now since we have three parallel lines (SR, AC, and PQ) with two transversals (PS and QR), the ratio of SA to AP must be the same as the ratio of RC to CQ. Lo and behold, that ratio is 1:1. So C is the midpoint of RQ. Thus the lines AC and AB (as you originally defined it) are the same line. So AB is parallel to PQ and SR. To get the second part of the problem, drop all these perpendiculars: P/----->---\Q /| |\ A/-H---------J-\B /| | | |\ S/---------->------\R D E F G Then since all the things that look like rectangles really are, you have the following equations: AB = PQ + DE + FG SD = AH (from congruent triangles) = DE GR = JB (from congruent triangles) = FG. So from the last two equations, D and G are the midpoints of SE and FR, respectively. So DE = SE/2, and FG = FR/2. So from the very first equation, AB = PQ + SE/2 + FG/2. And since SR = PQ + SE + FG, you can probably see that AB is the average of PQ and SR. Anyway, let us know if you didn't follow something, or if you think we messed up somewhere. -K |
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