Transcript Chapter 2
Chapter 2 EXISTENCE AND UNIQUENESS One of the basic axioms of Euclidean geometry says that two points determine a unique line. This implies that two distinct lines cannot intersect in two or more points, they can either intersect in only one point or not at all. Two lines that don’t intersect are called parallel. PROBLEM l Given a line l and a point P not on ‚ construct a line through P and parallel to l . L let A be any point on l, and draw AP Then draw a line PQ so that QPA PAB as shown in the figure. This will be the desired line. The proof will be by contradiction. If PQand l are not parallel, we may assume without loss of generality that they intersect as in the figure on the side of B at the point C. Now consider PAC. the exterior angle APQ is equal to the interior angle PAC. But this contradicts the exterior angle theorem, which states that QPA PAC . Hence PQ must be parallel to l . I Given a line l and a point P not on l , there exists a line that contains P and is parallel to l . COROLLARY Given lines ABand PQ as in the figure, if QPA PAB , then to PQ . AB is parallel THE PARALLEL POSTULATE If l is any line and P is a point not on l . then there is no more than one line through P parallel to l . Opposite Interior Angles Theorem Let AB and PQ be parallel lines with transversal AP such that QPA and PAB are opposite interior angles. Then QPA PAB . The proof will be by contradiction. If the theorem was false and if QPA PAB , then we could construct a distinct line PQ through P such that APQ PAB. Since APQ and PAB are opposite interior angles, their congruence implies that PQ AB . But this is now a contradiction of the parallel postulate : PQand PQare two different lines , each goes through P and each is parallel to AB. This contradiction comes about because we assumed that APQ PAB. So these angles must be congruent. THEOREM Let ABC be any triangle. then A B C 180 . Proof Let PAQbe the line through A parallel to BC such that B and BAPare opposite interior angles and C and CAQ are opposite interior angles, as in figure. so B BAP and C CAQ . Hence A B C A BAP CAQ =180. Since A , BAP and CAQ all together make a straight line. Let ABCD be any quadrilateral. then A B C D 360 We draw the diagonal AC thus breaking the quadrilateral into two triangles. Note that A B C D CAB CAD B ACB ACD D (ACD D DAC ) + (CAB B BCA). The first sum of the last expression represents the sum of the angles of ACD and the second sum represents the sum of the angles of CAB. Hence, each is 180 and together they add up to 360. COROLLARY(SAA) ABC and DEFassume that A D, B E and BC EF. In then ABC DEF. Given a quadrilateral ABCD, the following are equivalent: 1. AB CD and AD BC . 2. AB CDand AD BC. 3.The diagonals bisect each other. LEMMA l l Let be a line. P a point not on . And A and B distinct points on l such that PA is perpendicular to l .Then PA PB . L THEOREM l1 and l2 be parallel lines and let P and Q be points on l 2 Then the distance from P to l 1 equals the distance from Q to l1 Let Proof Draw lines from P and from Q perpendicular to l1, Meeting l1at B and at C , respectively. Since PBC 90 and QCB 90, these angles are congruent , Moreover . QBC is congruent to the supplement ofPBC. l2 l1 So PB QC By opposite interior angles. Similarly , PQ BC. Therefore PBCQ must be a parallelogram, since opposite sides are Parallel . Hence PB QC , as claimed.