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3.2 More Neutral Theorems Pasch’s Axiom-If a line l intersects Δ PQR at point S such that P-S-Q, then l intersects Crossbar Theorem-If X is a point in.

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Presentation on theme: "3.2 More Neutral Theorems Pasch’s Axiom-If a line l intersects Δ PQR at point S such that P-S-Q, then l intersects Crossbar Theorem-If X is a point in."— Presentation transcript:

1 3.2 More Neutral Theorems Pasch’s Axiom-If a line l intersects Δ PQR at point S such that P-S-Q, then l intersects Crossbar Theorem-If X is a point in the interior of triangle Δ UVW, then intersects at a point Y such that W-Y-V

2 3.3 Congruence Conditions We already had SAS congruence for triangles (postulate 15) Can also have ASA AAS SSS

3 Isosceles Triangle Theorem If 2 sides of a triangle are congruent then the angles opposite them are congruent. Converse Isosceles Theorem: If 2 angles of a triangle are congruent then the sides opposite them are congruent.

4 Converse Isosceles Proof Suppose we have ΔABC with

5 Inverse Isosceles Theorem If 2 sides of a triangle are not congruent, then the angles opposite those sides are not congruent, and the larger angle is opposite the larger side. Similarly, If two angles of a triangle are not congruent, then the sides opposite them are not congruent and the larger side is opposite the larger angle. Proof

6 Triangle Inequality Theorem Triangle Inequality Theorem: The sum of the measures of any 2 sides of a triangle is greater than the measure of the third. The Hinge Theorem: Draw picture

7 Do #6 and #10 Assign #4, #7

8 3.4 The Place of Parallels We have not talked about parallel lines because we have only assumed SMSG postulates 1-15 However, in Neutral Geometry it can be shown that at least one line can be drawn parallel to a given line through a point not on that line.

9 Alternate Interior Angle Thrm If two lines are intersected by a transversal such that a pair of alternate interior angles formed is congruent, then the lines are parallel. Read Proof Note: the converse (what you were probably given in high school geometry), which states that if two parallel lines are intersected by a transversal then the pairs of alternate interior angles are congruent is NOT a theorem in neutral geometry.

10 Corollary Two lines perpendicular to the same line are parallel. Proof: Given lines l, m, n such that l ┴ n, m ┴ n. Then wts: l is parallel to m. By definition, the intersection of perpendicular lines produces four right angles. Since l and m are both intersected by n at right angles, the alternate interior angles are congruent. By Alt. Interior Angle Thrm, l is parallel to m. Therefore two lines perpendicular to the same line are parallel.

11 Corollary If two lines are intersected by a transversal such that a pair of corresponding angles formed is congruent, then the lines are parallel. Proof: Given n is a transversal of l and m, and corresponding angles <2 and <6 are congruent. wts: l is parallel to m Since <2 and <4 are vertical angles, <2 ≡ <4. Since <2 ≡ <4 and <2 ≡ <6, by transitivity of congruence, <4 ≡ <6. Since <4 and <6 are alternate interior angles, by alt. int. angle thrm. we have that l is parallel to m.

12 Corollary If two lines are intersected by a transversal such that a pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel.

13 Proof Given n is a transversal of l and m and angles 1 and 6 are interior angles on the same side that are supplementary. wts: l is parallel to m Since <1 and <6 are supplementary, m<1+m<6 =180. We know that <6 and <7 are supplementary by definition, so m<6+m<7=180. This implies that m<1=m<7 and so <1 ≡ <7. Since <1 and <7 are congruent alt. interior angles, by alt. int. angle theorem, I is parallel to m

14 #3 Prove: If two lines are intersected by a transversal such that a pair of alternate interior angles is congruent, then the lines have a common perpendicular. Proof: Given n is a transversal of l and m and <1 ≡<7, and <4 ≡ <6. wts: l and m have common perp, i.e. let n be perpendicular to l then show n is perpendicular to m Since n is perp to l, then m<1=m<2=m<3=m<4=90 Since <1 ≡<7, and <4 ≡ <6, m<6=m<7=90. So m<8=m<5=90 since they are vertical angles. Therefore n is perpendicular to m. So they share a common perpendicular.

15

16 Equivalences to Euclidean Parallel Postulate Theorem (3.4.5) Euclid’s 5 th postulate is equivalent to the Euclidean parallel postulate. Euclid’s 5 th : That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

17 Proof (→) Assume Euclid’s 5 th and prove EPP. Given line m and point A not on m, draw line t perp to m through A which intersects m and B. Let l be a line through A perp to t. By Cor 3.4.2, l || m wts: there is not another line through A || to m Let n be another line through A || to m Since n ≠ l then m∩n = Ø and E is interior to

18 (←) Assume EPP and prove Euclid’s 5 th. Assume m<1 + m<2 < 180. Prove l intersects m on the same side of t. Draw ray AD such that

19 Other equivalences to EPP The EPP is equivalent to 1. The converse of the Alt. Int. Angle Theorem 2. If a line intersects one of two parallel lines, then it intersects the other. 3. If a line is perpendicular to one of two parallel lines, then it is perp to the other. 4. Given ΔPQR and any line segment AB, there exists ΔABC having a side ≡ to AB and ΔABC is similar to ΔPQR.

20 Assign #5, 6, 10, 11 #5 and #10 for turn in (due Monday Sep 24)

21 3.5 The Saccheri-Legendre Theorem Saccheri studied a specific class of quadrilaterals now called Saccheri quadrilaterals. םABCD where AB ≡ CD,

22 Saccheri-Legendre Theorem: The angles sum of any triangle is less than or equal to 180 Lemma The sum of any 2 angles of a triangle is less than 180 Lemma For any ΔABC, there exists ΔA’B’C’ having the same angle sum but m

23 Cor The angle sum of any convex quadrilateral is less than or equal to 360 Convex: םABCD is convex if a pair of opposite sides exist AB and CD such that CD is contained in one of the half-planes bounded by line AB and AB is in one of the half planes bounded by line CD

24 Proof wts: m

25 #3 State and prove the converse of Euclid’s 5 th. If 2 straight lines meet on a particular side of a transversal, then the sum of the interior angle on that side is less than 2 right angles (180)

26 Proof Let l and m be lines such that l∩m = A and let t be a transversal of I and m. wts: <1 + <2 < 180 Let B = l∩t and C = t∩m and consider ΔABD By lemma m

27 From this Saccheri easily showed that in the quadrilateral

28 3.6 The Search for a Rectangle For SQ םABCD: AD, BC called sides or legs AB called base DC called summit

29 Theorem The diagonals of a Saccheri quadrilateral are congruent Proof: Let םABCD be a SQ. So then AD ≡ BC,

30 Theorem The summit angles of a SQ are congruent Theorem The summit angles of a SQ are not obtuse, thus are both acute or both right Theorem Trying to prove two quadrilaterals are congruent, need to show SASAS

31 Theorem The line joining the midpoints of both the summit and the base of a SQ is perpendicular to both. Proof: Given SQ םABCD, let MN join the midpoints of AB and DC wts: MN ┴ AB and MN ┴ CD By def, AD ≡ BC and

32 By def of MN, AM=MB and DN=NC So םAMND ≡ םBMNC by SASAS theorem This implies

33 Cor The summit and base of a SQ are parallel Theorem In any SQ, the length of the summit is greater than or equal to the length of the base.

34 Lambert Lambert Quadrilateral: quadrilateral with at least 3 right angles Lambert’s Theorem The fourth angle of a LQ is not obtuse, thus is either acute or right.

35 Proof Let םPQRS be a LQ Then


36 Theorem The measure of the side included between 2 right angles of a LQ is less than or equal to the measure of the side opposite it. Proof: Let םPQRS be a LQ. Then SR and choose P’ such that P’Q = SR. Then םP’QRS is a SQ So then <1 ≡ <2 and m<1 = m<2 ≤ 90 This is a contradiction by exterior angle theorem to ΔPP’S So PQ ≤ SR Similarly QR ≤ PS (you try to finish this one)

37 Theorem The measure of the line joining the midpoints of the base and summit of at SQ is less than or equal to the measure of it sides. Proof: in book Lots of theorems and corollaries about when triangles and rectangles exist that you should look at

38 Theorem: The EPP is equivalent to saying the angle sum of every triangle is 180 Theorem: The Pythagorean Theorem implies the existence of a rectangle. Theorem: The Pythagorean Theorem is equivalent to EPP

39 #25 If a quadrilateral is both SQ and LQ then it is a rectangle. Proof: Given םABCD such that AB ≡ BC and


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