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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 the interior of triangle Δ UVW, then intersects at a point Y such that W-Y-V

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**3.3 Congruence Conditions**

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

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**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.

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**Converse Isosceles Proof**

Suppose we have ΔABC with <A ≡ <C. Now draw as the bisector of <ABC. Let D be the intersection of with Want 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.

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**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

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**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

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

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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.

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**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.

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**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.

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Corollary 3.4.3 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.

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Corollary 3.4.4 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.

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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

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#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.

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**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.

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**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 || 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 <DAB So 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.

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**(←) 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 ≠ Ø In 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 < <3 This is a contradiction, so l∩m on the RHS.

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**Other equivalences to EPP**

The EPP is equivalent to The converse of the Alt. Int. Angle Theorem If 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.

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

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**3.5 The Saccheri-Legendre Theorem**

Saccheri studied a specific class of quadrilaterals now called Saccheri quadrilaterals. םABCD where AB ≡ CD, <A ≡ <D =90 So either <B and <C are right angles, acute angles or obtuse angles He tried to show that if they were acute/obtuse you get contradictions

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**Saccheri-Legendre Theorem:**

The angles sum of any triangle is less than or equal to 180 Lemma 3.5.2 The sum of any 2 angles of a triangle is less than 180 Lemma 3.5.3 For any ΔABC, there exists ΔA’B’C’ having the same angle sum but m<A’ ≤ ½ m<A To prove theorem, reused this lemma repeatedly

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Cor 3.5.4 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

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**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 ΔBDC Thrm says sΔABD ≤ 180 and sΔBDC ≤ 180 So sΔABD + sΔBCD ≤ 360 From angle addition postulate <ADB + <BDC=<D and <ABD + <CBD =<B So m<A + m<B +m<C + m<D ≤ 360

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**#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)

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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<C +m<B < 180 So the sum of the interior angles on the same side of a transversal is < 180.

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**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

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**3.6 The Search for a Rectangle**

For SQ םABCD: AD, BC called sides or legs AB called base DC called summit <C, <D called summit angles His goal: show <C=<D=90 So he looked at ways to maybe construct a rectangle in neutral geometry rectangle: quadrilateral with 4 right angles

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**Theorem 3.6.1 The diagonals of a Saccheri quadrilateral are congruent**

Proof: Let םABCD be a SQ. So then AD ≡ BC, <A ≡ <B =90 Draw diagonals AC and DB and consider ΔABC and ΔBAD Since AB ≡ AB, and <A ≡ <B, AD ≡ BC Then ΔABC ≡ ΔBAD by SAS So DB ≡ AC.

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**Theorem 3.6.2 Theorem 3.6.3 Theorem 3.3.4**

The summit angles of a SQ are congruent Theorem 3.6.3 The summit angles of a SQ are not obtuse, thus are both acute or both right Theorem 3.3.4 Trying to prove two quadrilaterals are congruent, need to show SASAS

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Theorem 3.6.4 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 <A ≡ <B =90 Also, <C ≡ <D by theorem 3.6.2

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**By def of MN, AM=MB and DN=NC**

So םAMND ≡ םBMNC by SASAS theorem This implies <NMB ≡ <NMA or m<NMB=m<NMA We know m<NMA + m<NMB =180 So 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

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**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.

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Lambert Lambert Quadrilateral: quadrilateral with at least 3 right angles Lambert’s Theorem 3.6.7 The fourth angle of a LQ is not obtuse, thus is either acute or right.

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**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 =270 Then <S ≤ 90 So it is either acute or right.

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Theorem 3.6.8 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 <P=<Q=<R=90 and <S ≤ 90 wts: PQ ≤ SR and QR ≤ PS Assume PQ > 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)

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Theorem 3.6.9 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

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

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**#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=90 wts: םABCD is a rectangle. By SQ theorem, <C ≡<D So then <D=90 So םABCD is a rectangle. Homework: #1, 3, 4, 6, 7, 12, 14, 26 Turn in #3, 6, 14

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Sect. 4.6 Isosceles, Equilateral, and Right Triangles

Sect. 4.6 Isosceles, Equilateral, and Right Triangles

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