1.1. Some constructions 3 radius r that lies inside ∠AOB. See Figure 1.1.2. Denote the points where this arc intersects the segments OA and OB by P and Q, respectively. Using Figure 1.1.2 a radius s different from r and satisfying s 1 2 dist (P, Q), draw the circle with center P . Now draw the circle with center Q and radius s. (Why did we choose s 1 2 dist (P, Q)?) Denote by R the point of intersection of these two circles that lies inside the angle. Now use the straightedge to draw the line segments OR, PR, and QR. See Figure 1.1.3. Figure 1.1.3 The claim is that the line OR bisects ∠AOB. To see this we prove that the triangles ΔOPR and ΔOQR are congruent. Indeed, first note that |PR| = |QR| = s. Also |OP| = |OQ| = r. Since the remaining side, OR, is held in common by the triangles, ΔOPR ∼ ΔOQR. Therefore OR bisects ∠AOB. 1.1.2. Corollary. If a line is given as well as a point P on the line, using straightedge and compass alone it is possible to construct a straight line that is perpendicular to and intersects at the point P . Proof. Put points A and B on , one on each side of P . Now use the preceding proposition to bisect ∠APB. The next result is similar to the preceding one with one major difference. 1.1.3. Proposition. Given a straight line and a point Q not on this line, using straightedge and compass alone it is possible to construct a straight line passing through Q and perpendicular to .

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