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 TheoremsPasch’s Axiom-If a line l intersects Δ PQR at point S such that P-S-Q, then l intersectsCrossbar 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 haveASAAASSSS
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 <A ≡ <C. Now draw as the bisector of <ABC. Let D be the intersection of withWant to Show that ΔABD ≡ ΔCBD.Since <BAC ≡ <BCA, <ABD ≡ <CBD, and ≡ , we know that ΔABD ≡ ΔCBD by AAS, therefore ≡Hence, the sides opposite the angles are congruent.
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 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
8 3.4 The Place of ParallelsWe have not talked about parallel lines because we have only assumed SMSG postulates 1-15However, 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 ProofNote: 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 3.4.2 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 3.4.3If 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 mSince <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 3.4.4If 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 ProofGiven 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 mSince <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 #3Prove: 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 mSince n is perp to l, then m<1=m<2=m<3=m<4=90Since <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.
16 Equivalences to Euclidean Parallel Postulate Theorem (3.4.5) Euclid’s 5th postulate is equivalent to the Euclidean parallel postulate.Euclid’s 5th: 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 5th 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 || mwts: there is not another line through A || to mLet n be another line through A || to mSince n ≠ l then m∩n = Ø and E is interior to <DABSo m<EAB < m<DAB and so <EAB is acute.Therefore m<EAB + m<ABC < 180 and m∩n ≠ Ø which is a contradiction.So there is only one parallel line through A.
18 (←) Assume EPP and prove Euclid’s 5th. Assume m<1 + m<2 < 180.Prove l intersects m on the same side of t.Draw ray AD such that <BAD ≡ <3.So ray AD || m (Alt. Int. Angle Thrm)So since l ≠ AD, the EPP guarantees l∩m ≠ ØIn order to show I and m meet on the same side of m (on the RHS), we will assume l∩m on the LHS of t.Then <1 is exterior angle of ΔABC.This implies <1 < <DAB which implies <1 < <3This is a contradiction, so l∩m on the RHS.
19 Other equivalences to EPP The EPP is equivalent toThe converse of the Alt. Int. Angle TheoremIf a line intersects one of two parallel lines, then it intersects the other.If a line is perpendicular to one of two parallel lines, then it is perp to the other.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, <A ≡ <D =90So either <B and <C are right angles, acute angles or obtuse anglesHe tried to show that if they were acute/obtuse you get contradictions
22 Saccheri-Legendre Theorem: The angles sum of any triangle is less than or equal to 180Lemma 3.5.2The sum of any 2 angles of a triangle is less than 180Lemma 3.5.3For any ΔABC, there exists ΔA’B’C’ having the same angle sum but m<A’ ≤ ½ m<ATo prove theorem, reused this lemma repeatedly
23 Cor 3.5.4The angle sum of any convex quadrilateral is less than or equal to 360Convex: 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<A + m<B +m<C + m<D ≤ 360 Assume we have convex םABCD.Draw line segment BD.It separates םABCD into ΔABD and ΔBDCThrm says sΔABD ≤ 180 and sΔBDC ≤ 180So sΔABD + sΔBCD ≤ 360From angle addition postulate <ADB + <BDC=<D and <ABD + <CBD =<BSo m<A + m<B +m<C + m<D ≤ 360
25 #3 State and prove the converse of Euclid’s 5th. 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 ProofLet l and m be lines such that l∩m = A and let t be a transversal of I and m.wts: <1 + <2 < 180Let B = l∩t and C = t∩m and consider ΔABDBy lemma m<C +m<B < 180So the sum of the interior angles on the same side of a transversal is < 180.
27 From this Saccheri easily showed that in the quadrilateral <B and <C were not obtuse A contradiction to acute angles still hasn’t been found.Homework: #2
28 3.6 The Search for a Rectangle For SQ םABCD:AD, BC called sides or legsAB called baseDC called summit<C, <D called summit anglesHis goal: show <C=<D=90So he looked at ways to maybe construct a rectangle in neutral geometryrectangle: quadrilateral with 4 right angles
29 Theorem 3.6.1 The diagonals of a Saccheri quadrilateral are congruent Proof:Let םABCD be a SQ. So then AD ≡ BC, <A ≡ <B =90Draw diagonals AC and DB and consider ΔABC and ΔBADSince AB ≡ AB, and <A ≡ <B, AD ≡ BCThen ΔABC ≡ ΔBAD by SASSo DB ≡ AC.
30 Theorem 3.6.2 Theorem 3.6.3 Theorem 3.3.4 The summit angles of a SQ are congruentTheorem 3.6.3The summit angles of a SQ are not obtuse, thus are both acute or both rightTheorem 3.3.4Trying to prove two quadrilaterals are congruent, need to show SASAS
31 Theorem 3.6.4The 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 DCwts: MN ┴ AB and MN ┴ CDBy def, AD ≡ BC and <A ≡ <B =90Also, <C ≡ <D by theorem 3.6.2
32 By def of MN, AM=MB and DN=NC So םAMND ≡ םBMNC by SASAS theoremThis implies <NMB ≡ <NMA or m<NMB=m<NMAWe know m<NMA + m<NMB =180So then m<NMB =½(180) and m<NMA=½(180)Similarly, m<MND=½(180) and m<MNC=½(180)Therefore, MN ┴ AB and MN ┴ CD
33 Cor. 3.6.5 Theorem 3.6.6 The summit and base of a SQ are parallel In any SQ, the length of the summit is greater than or equal to the length of the base.
34 LambertLambert Quadrilateral: quadrilateral with at least 3 right anglesLambert’s Theorem 3.6.7The fourth angle of a LQ is not obtuse, thus is either acute or right.
35 Proof Let םPQRS be a LQ Then <P = <Q = <R =90 wts: <S ≤ 90 Since <P + <Q + <R + <S ≤ 360 by Cor and <P + <Q + <R =270Then <S ≤ 90So it is either acute or right.
36 Theorem 3.6.8The 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 <P=<Q=<R=90 and <S ≤ 90wts: PQ ≤ SR and QR ≤ PSAssume PQ > SR and choose P’ such that P’Q = SR.Then םP’QRS is a SQSo then <1 ≡ <2 and m<1 = m<2 ≤ 90This is a contradiction by exterior angle theorem to ΔPP’SSo PQ ≤ SRSimilarly QR ≤ PS (you try to finish this one)
37 Theorem 3.6.9The 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 bookLots 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 180The Pythagorean Theorem implies the existence of a rectangle.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 <A=<B=<C=90wts: םABCD is a rectangle.By SQ theorem, <C ≡<DSo then <D=90So םABCD is a rectangle.Homework: #1, 3, 4, 6, 7, 12, 14, 26Turn in #3, 6, 14