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Feb 23 Notes: Definition: Two Lines L and M Are Parallel If They Lie in The Feb 23 Notes: Definition: Two lines l and m are parallel if they lie in the same plane and do not intersect. Terminology: When one line intersects each of two given lines, we call that line a transversal. We define alternate interior angles, corresponding angles, alternate exterior angles, and interior angles on the same side of the transversal using various betweeness and half-plane notions. Suppose line l intersects lines m and n at points B and E, respectively, with points A and C on line m and points D and F on line n such that A-B-C and D-E-F, with A and D on the same side of l. Suppose also that G and H are points such that H-E-B- G. Then pABE and pBEF are alternate interior angles, as are pCBE and pDEB. pABG and pFEH are alternate exterior angles, as are pCBG and pDEH. pGBC and pBEF are a pair of corresponding angles, as are pGBA & pBED, pCBE & pFEH, and pABE & pDEH. pCBE and pFEB are interior angles on the same side of the transversal, as are pABE and pDEB. Our Last Theorem in Absolute Geometry: If two lines in the same plane are cut by a transversal so that a pair of alternate interior angles are congruent, the lines are parallel. Proof: Let l intersect lines m and n at points A and B respectively. Let p1 p2. Suppose m and n meet at point C. Then either p1 is exterior to ªABC, or p2 is exterior to ªABC. In the first case, the exterior angle inequality gives p1 > p2; in the second, it gives p2 > p1. In either case, we have a contradiction to p1 p2. The converse of this statement is one way of stating the Euclidean Parallel Postulate. There are many other formulations as well. Here is a list of a few statements equivalent to Parallel Postulate. 1. If two parallel lines are cut by a transversal, alternate interior angles are congruent. 2. If two lines in the same plane are cut by a transversal so that the sum of the measures of a pair of interior angles on the same side of the transversal is less than 180, then the lines will meet on that side of the transversal. 3. A third line intersecting one of two parallel lines intersects the other. 4. A line perpendicular to one of two parallel lines is perpendicular to the other. 5. The perpendicular bisectors of the sides of a triangle are concurrent. 6. There exists a circle passing through any three noncollinear points. 7. The sum of measures of the angles of a triangle is 180. 8. There exists one triangle such that the sum of measures of the angles is 180. 9. There exists a rectangle. 10. There exist two lines l and m such that l is equidistant from m. 11. There exists a pair of similar, noncongruent triangles. 12. There exists an acute angle such that every line intersecting and perpendicular to one ray of the angle intersects the other ray. We take as our version of the Parallel Postulate the following: Axiom P-1: If l is any line and P any point not on l, there exists in the plane of l and P one and only one line m that passes through P and is parallel to l. We can now prove Statement 1 on our list, namely: Theorem: If two parallel lines are cut by a transversal, alternate interior angles are congruent. Proof: Let l and m be parallel lines cut by a transversal t at points Q and P, respectively. Suppose for contradiction that alternate interior angles p1 (at point P) and p2 (at point Q) are not congruent, and WLOG that mp1 > mp2. Using the angle construction theorem, create ray on the other side of t from p2 such that mpRPQ = mp2. Then from the above theorem, since pRPQ and p2 are alternate interior angles, it follows that lines l and are parallel. But that means there are two lines through P parallel to line l, contradicting the parallel postulate. So p1 and p2 are congruent. Now comes a flood of straightforward corollaries. First, notice that, given two lines in the same plane cut by a transversal, The following are equivalent: 1. A pair of alternate interior angles are congruent 2. Both pairs of alternate interior angles are congruent. 3. Both pairs of alternate exterior angles are congruent 4. All pairs of corresponding angles are congruent. 5. Both pairs of interior angles lying on one side of the transversal are supplementary. This is very easy to prove using vertical angles and linear pairs so we’ll leave the proof as an exercise. From these equivalences come the C, F, and Z properties listed in your text on page 215. Taken together with the above two theorems, we have the following: If two lines in the same plane are cut by a transversal, then the lines are parallel iff • alternate interior angles are congruent • alternate exterior angles are congruent • corresponding angles are supplementary • interior angles on the same side of the transversal are supplementary • exterior angles on the same side of the transversal are supplementary An immediate consequence is: If a line is perpendicular to one of two parallel lines, it is perpendicular to the other also. Euclidean Exterior Angle Theorem: In any triangle, the measure of an exterior angle is the sum of the measures of the two remote interior angles. Proof: We use the same construction as for the proofs of the exterior angle theorem and Saccheri - Legendre Theorem in absolute geometry. Begin with ªABC and point D with B-C-D. Find the midpoint M of and find point E with B-M-E and MB = ME. Connect E and C. Now by construction and SAS, ªBAM is congruent to ªECM, and it is easy to show that E is interior to pACD. Moreover, since pBAC pECA, with as the transversal we have congruent alternate interior angles and thus . This makes pABC and pECD corresponding angles (with as the transversal), and hence congruent. Thus Another flood of corollaries: • The sum of the measure of the angles of any triangle is 180. • The acute angles of a right triangle are complementary • The sum of the measures of any convex quadrilateral is 360 • Rectangles exist; in fact, every Saccheri quadrilateral and every Lambert quadrilateral is a rectangle • Squares exist. This sharpens most of the “almost Euclidean” results we’ve gotten in the last few chapters. We have one left: Midpoint Connector Theorem: The segment joining the midpoints of two sides of a triangle is parallel to the third side and has length one-half the third side. Proof: Begin with ª ABC with L and M the midpoints of and respectively. Extend to find P with L-M-P and LM = MP. Connect P and C. By construction and SAS, ªLAM ªPCM and so (pLAC pPCA, alternate interior angles). Note also that BL = LA = PC. If we construct , we have pBLC pPCL (alternate interior angles with as the transversal) so we get ªBLC ªPCL by SAS. Thus by CPCF, and since pCLP pLCB, and they are alternate interior angles with as the transversal, . Corollary: If a line bisects one side of a triangle and is parallel to the second, it also bisects the third side. (Follows from uniqueness of parallels.) .
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