University:
California State University, Northridge
Course:
MATH 310 | Basic Concepts of Geometry
Academic year:

2006
Views:

353

Pages:

1
Author:

Jeremiah Sutton

---£------�H----- Given: ABCDEF is an equiangular hexagon Prove: AB+BC = FE+ED I I -ft-, I A, , , __,_ x, I - - ti- - 13 ri C I. ABCDEF is equiangular hexagon (Given) 2. All angles arc equal to 120 degrees (interior angles of an equiangular hexagon are 120 degrees) 3. Extend AF towards A and extend BC towards B so the two rays intersect at X. (Construction) 4. Extend FE towards E and extend DC towards D so the two rays intersect at Y. (Construction) 5. Angle XAB and XBA are 60 degrees (exterior angles of an equiangular hexagon are 60 degrees) 6. Angle YED and YDE are 60 degrees (exterior angles ofan equiangular hexagon are 60 degrees) 7. Angle X and Y are 60 degrees ( triangle sum theorem) 8. 6. XAB and 6. YEO are equilateral triangles (definition of equilateral triangles ) 9. XB = AB and YE = ED ( definition of equilateral triangles) I 0. Angles C and F arc equal (Given) 11. XFYC is Parallelogram (Ifboth pairs ofopposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram) 12. FY= XC (Ifa quadrilateral is a parallelogram, then the opposite sides are congruent) 13. FY= FE+ EY; XC = XB + BC ( Segment addition posllllate) 14. FE+ EY= XB + BC ( substitution) 15. FE+ ED= AB + BC ( substitution)