SOLUTION: I need help with a proof. It is a pentagon with a star within it, the letters on the outside of the pentagon starting at the top and going righ are D, C, B, A and E. The only lette

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Question 173746This question is from textbook Geometry
: I need help with a proof. It is a pentagon with a star within it, the letters on the outside of the pentagon starting at the top and going righ are D, C, B, A and E. The only letter on the interior is an O in the bottom of a smaller pentagon formed by the star which is in the middle of the original. My goal is to make EC paralell to AB, please, please help if you can. This question is from textbook Geometry

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!


1. The sum of the interior        (n-2)*180° = (5-2)*180° =
angles is 540°                    3*180° = 540°

2. ÐABC = 72°                     All interior angles of a regular 
                                  polygon are equal, and 540°÷5=72°

3. AB = BC                        All sides of a regular polygon 
                                  are equal. 

4. DABC is isosceles              Two sides equal, AB = BC      
                               
5. ÐBAC+ÐBCA+ÐABC=180°            The sum of the interior angles
                                  of a triangle is 180°

6. ÐBAC+ÐBCA+108°=180°            Substituting 72° for ÐABC, since
                                  they are equal.

7. ÐBAC+ÐBCA=72°                  Subtracting equals from equals,
                                  (subtract 108° from both sides)

8. ÐBAC = ÐBCA                    Base angles of isoceles DABC

9. ÐBAC = ÐBCA = 36°              Equal angles, each half of 72°, from 7


10. ÐCDE = 72°                     All interior angles of a regular 
                                   polygon are equal, and 540°÷5=72°

11. CD = DE                        All sides of a regular polygon 
                                   are equal. 

12. DCDE is isosceles              Two sides equal, CD = DE      
                               
13. ÐDCE+ÐDEC+ÐCDE=180°            The sum of the interior angles
                                   of a triangle is 180°

14. ÐDCE+ÐDEC+108°=180°           Substituting 72° for ÐCDE, since
                                  they are equal.

15. ÐDCE+ÐDEC=72°                 Subtracting equals from equals,
                                  (subtract 108° from both sides)

16. ÐDCE = ÐDEC                    Base angles of isoceles triangle CDE

17. ÐDCE = ÐDEC = 36°              Equal angles, each half of 72°, from 15.

18. ÐBCA = 36°                     From 9.

19. ÐDCE = 36°                     From 17.

20. ÐBCA+ÐACE+ÐDCE = ÐBCD          Whole = sum of parts.

21. ÐBCD = 108°                    Reason 2

22. ÐBCA+ÐACE+ÐDCE = 108°          Substituting equals for equals

23. 36° + ÐACE + 36° = 108°        Subs. equals for equals, using
                                   18 and 19.

24. 72° + ÐACE = 108°              Adding 36°+36° in 22.

25. ÐACE = 36°                     Subtract equals from equals.
                                   (Subtract 72° from both sides of 25.
26. ÐBAC = 36°                     Step 9

27. ÐACE = ÐBAC                    Both equal 36°

28. EC ú÷ AB                       Equal alternate interior angles ÐACE and
                                   ÐBAC when transversal AC cuts lines
                                   EC and AB
Edwin
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