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3.3 Parallel Lines & Transversals

3.3 Parallel Lines & Transversals. Geometry Presented by Mrs. Spitz Fall 2005. Standard/Objectives:. Standard 3: Students will learn and apply geometric concepts. Objectives: Prove and use results about parallel lines and transversals. Use properties of parallel lines. Assignment.

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3.3 Parallel Lines & Transversals

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  1. 3.3 Parallel Lines & Transversals Geometry Presented by Mrs. Spitz Fall 2005

  2. Standard/Objectives: Standard 3: Students will learn and apply geometric concepts. Objectives: • Prove and use results about parallel lines and transversals. • Use properties of parallel lines.

  3. Assignment • pp. 146 -147 #1-27 • Quiz after this section next class meeting . . . don’t miss it. There is another quiz after 3.5.

  4. Post. 15 – Corresponding s Post. • If 2  lines are cut by a transversal, then the pairs of corresponding s are . • i.e. If l m, then 12. 1 2 l m

  5. Thm 3.4 – Alt. Int. s Thm. • If 2  lines are cut by a transversal, then the pairs of alternate interior s are . • i.e. If l m, then 12. 1 2 l m

  6. Statements l m 3  2 1  3 1  2 Reasons Given Corresponding s post. Vert. s Thm   is transitive Proof of Alt. Int. s Thm. 3 1 2 l m

  7. Thm 3.5 – Consecutive Int. s thm • If 2  lines are cut by a transversal, then the pairs of consecutive int. s are supplementary. • i.e. If l m, then 1 & 2 are supp. l m 1 2

  8. Thm 3.6 – Alt. Ext. s Thm. • If 2  lines are cut by a transversal, then the pairs of alternate exterior s are . • i.e. If l m, then 12. l m 1 2

  9. Thm 3.7 -  Transversal Thm. • If a transversal is  to one of 2  lines, then it is  to the other. • i.e. If l m, & t  l, then t m. ** 1 & 2 added for proof purposes. t 1 2 l m

  10. Statements l m, t  l 12 m1=m2 1 is a rt.  m1=90o 90o=m2 2 is a rt.  t m Reasons Given Corresp. s post. Def of  s Def of  lines Def of rt.  Substitution prop = Def of rt.  Def of  lines Proof of  transversal thm

  11. Ex: Find: m1= m2= m3= m4= m5= m6= x= 1 125o 2 3 5 4 6 x+15o

  12. Ex: Find: m1=55° m2=125° m3=55° m4=125° m5=55° m6=125° x=40° 1 125o 2 3 5 4 6 x+15o

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