Question

Given: In ∆ACB, c2 = a2 + b2.
Prove: ∆ACB is a right angle.

Complete the flow chart proof with missing reasons to prove that ∆ACB is a right angle.

Top path: by Construction, angle DFE is a right angle; angle ACB is a right angle. Next path, by Construction, line segment DF is congruent to line segment AC; by Labeling, let line segment AC equal b, line segment CB equal a, and line segment AB equal c; by Substitution, f-squared equals a-squared plus b-squared; by Transitive Property of Equality, f-squared equals c-squared; by Square root Property of Equality, f equals c; by SSS Postulate

Answer 1

Top path: by Construction, angle DFE is a right angle; angle ACB is a right angle. Next path, by Construction, line segment DF is congruent to line segment AC; by Labeling, let line segment AC equal b, line segment CB equal a, and line segment AB equal c; by Substitution, f-squared equals a-squared plus b-squared; by Transitive Property of Equality, f-squared equals c-squared; by Square root Property of Equality, f equals c; by SSS Postulate triangle ACB is congruent to triangle DFE; by CPCTC, angle ACB is a right angle. Next path, by Construction, line segment DF is congruent to line segment AC; by Labeling, let line segment DF equal e, line segment FE equal d, and line segment DE equal f; by Pythagorean Theorem, f-squared equals d-squared plus e-squared; by Substitution, f-squared equals a-squared plus b-squared; by Transitive Property of Equality, f-squared equals c-squared; by Square root Property of Equality, f equals c; by SSS Postulate triangle ACB is congruent to triangle DFE; by CPCTC, angle ACB is a right angle. Next path, by Construction, line segment FE is congruent to line segment CB; by Labeling, let line segment AC equal b, line segment CB equal a, and line segment AB equal c; by Substitution, f-squared equals a-squared plus b-squared; by Transitive Property of Equality, f-squared equals c-squared; by Square root Property of Equality, f equals c; by SSS Postulate triangle ACB is congruent to triangle DFE; by CPCTC, angle ACB is a right angle. Next path, by Construction, line segment FE is congruent to line segment CB; by Labeling, let line segment DF equal e, line segment FE equal d, and line segment DE equal f; by Pythagorean Theorem, f-squared equals d-squared plus e-squared; by Substitution, f-squared equals a-squared plus b-squared; by Transitive Property of Equality, f-squared equals c-squared; by Square root Property of Equality, f equals c; by SSS Postulate triangle ACB is congruent to triangle DFE; by CPCTC, angle ACB is a right angle. Next path, by Construction, Draw line segment ED; let line segment DF equal e, line segment FE equal d, and line segment DE equal f; by Pythagorean Theorem, f-squared equals d-squared plus e-squared; by Substitution, f-squared equals a-squared plus b-squared; by Transitive Property of Equality, f-squared equals c-squared; by Square root Property of Equality, f equals c; by SSS Postulate triangle ACB is congruent to triangle DFE; by CPCTC, angle ACB is a right angle. Bottom path, by Given c-squared equals a-squared plus b-squared; by Transitive Property of Equality, f-squared equals c-squared; by Square root Property of Equality, f equals c; by SSS Postulate triangle ACB is congruent to triangle DFE; by CPCTC, angle ACB is a right angle.

Which pair of reasons correctly completes this proof?

Answer 2

:O oh. my.

Answer 3

told you!

Answer 4

well....should i just close it now?

Answer 5

haha yeah. i tried reading it.. i'm so confused. i'm sorry :(

Answer 6

haha its ok...ill post a shorter one!!

Answer 7

do you know what a flow chart is?

Answer 8

mhmm

Answer 9

so you're supposed to put these things in order? or what?

Answer 10

i have no clue...

Answer 11

:| well, thats confusing. okay. post the shorter one.