Question 965887: Prove that a diagonal of a rhombus bisects each vertex angle through which it passes
Given: ABCD is a rhombus, AC is a diagonal
Statement: line AB in congruent to line BC
Angle 1 is congruent to angle 3
Line BC is II AC, AB is II to CD
<2 in congruent <3, <1 is congruent to <4
<1 is congruent to <2, <3 is congruent to <4
AC bisects
I need help with the reasons for my statement, thank you
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
Given: Rhombus ABCD with diagonal AC
To prove: ACF bisects ∠BAD and ∠BCD
1. AB ≅ CD 1. opposite sides of a parallelogram are congruent
2. BC ≅ AD 2. opposite sides of a parallelogram are congruent
3. AC ≅ AC 3. a line is congruent to itself.
4. ΔABC ≅ ΔADC 4. SSS 1,2,3
5. ∠1 ≅ ∠4 5. corresponding parts of congruent triangles.
6. ∠3 ≅ ∠2 6. corresponding parts of congruent triangles.
7. AB ≅ BC 7. definition of rhombus, all sides congruent.
8. ΔABC is isosceles 8. two sides congruent, 7
9. ∠1 ≅ ∠3 9. base angles of an isosceles triangle
10. ∠1 ≅ ∠2 10. things congruent to the same thing are equal
to each other, 6,9 11. AC bisects ∠ABD 11. from 10
12. ∠3 ≅ ∠4 12. things congruent to the same thing are equal
to each other, 5,9
13. AC bisects ∠BCD 13. from 12
Edwin
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