4 1. Trisecting Angles Proof. Choose a radius r for the compass that is larger than the distance from Q to and draw the circle with center Q and radius r. Let A and B be the points where this circle intersects . Draw the segments QA and QB. See Figure 1.1.4. Figure 1.1.4 Using A as the center, draw the circle with this same radius r. Similarly draw the circle with center B and radius r. Note that these circles intersect at the point Q as well as another point that lies on the side of opposite to that of Q call this second point where the circles meet P . See Figure 1.1.5. Draw the line segments AQ, BQ, AP , BP as in Figure 1.1.5. Figure 1.1.5 We want to show that PQ ⊥ . If I denotes the point where PQ and intersect, this means we want to show that ∠BIQ is a right angle. First note that |QA| = |QB| = r and |AP| = |BP| = r. Since the triangles ΔAQP and ΔBQP share the side PQ, this implies that ΔAQP ∼ ΔBQP . Hence ∠IQA = ∠IQB. So ΔIQA and ΔIQB have two pairs of sides equal and the corresponding angles that these pairs of sides form is equal. Therefore ΔIQA ∼ ΔIQB. Thus ∠QIA = ∠QIB. Since these two angles sum to the straight angle ∠BIA, it must be that each of them is a right angle. 1.1.4. Corollary. Given a straight line and a point Q not on , using straightedge and compass alone it is possible to construct a line passing through Q that is parallel to . Proof. Use the preceding proposition to construct a line passing through Q and perpendicular to . Let P be the point where this per- pendicular meets . Now use Corollary 1.1.2 to construct a line through Q and perpendicular to PQ. This line is parallel to . See Figure 1.1.6

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