Section 8.1 217 3. (a) [BB] Let A be any point on f and let r = IAPI. Draw the circle with center P and radius r. If this meets f in only the single point A, then PA is the desired perpendicular to f. Otherwise, the circle meets f in two points, A and B. Draw arcs with centers A and B and radius r meeting in P 4. and Q. Then PQ is the desired perpendicular to f. '\Pi To see this, consider the diagram at the right, with M the point of intersection of PQ and f. Since triangles P AQ and P BQ are .........4-congruent, LAPM = LBPM. Then it follows that .6.APM == .6.BP M and thus, LAMP = LBMP. Since these angles have sum 1800, each is a right angle. I.J .........f (b) Draw an arc with center P and any radius r and let A and B be the points of intersection of this arc with f. Then draw arcs with centers A and B and any radius s > r. Letting C be one of the points of intersection of these \r:l arcs, CP is the required perpendicular. To see this, note that triangles CPA and C P B have three pairs of equal sides and so are congruent. Thus, y1 .fJ' f LC P A = LC P B
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