Unformatted text preview: AB, choose a radius r (for example, r = lAB!) such that the arcs with centers A and B and radius r meet at two distinct points, P and Q. The point of intersection, M, of AB and PQ is the required midpoint. To see why this works, consider the diagram to the right, in which the labels are given as before and line segments AP, BP, AQ, BQ are joined. Note that LP AB = LP BA since 6.AP B is isosceles. Since D.BPQ is congruent to 6.APQ (three pairs of sides of equal length), LBPQ = LAPQ. Thus, triangles PAM and P BM are congruent (two equal pairs of sides and equal contained angles), so IAMI = IBMI as required. 2. Let A be any point of £. Draw an arc with center A and radius r = IAPI. Let B be the intersection of this arc with t. Draw arcs with radius r and centers P and B meeting in A and C. Then the line through PC is the desired line. To see this, note that PABC has four equal sides, so it's a parallelogram. A ~+t=+7 B...
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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