QUESTION IMAGE
Question
- complete the proof below by choosing the missing statements and justifications.
given: ( overline{ac} cong overline{df} ), ( overline{bc} cong overline{de} ). prove: ( overline{ab} cong overline{ef} )
(there is a diagram with segments: ( a---b---c ) and ( d---e---f ))
| statements | justification |
|---|---|
| ( overline{bc} cong overline{de} ) | given |
statements + justification bank:
- definition of congruence
- ( bc = de )
- ( ab + bc = ac )
- ( ac = df )
- segment addition postulate
- definition of congruence
- given
- given
- ( de + ef = df )
- ( ab cong ef )
- segment addition postulate
- ( ab cong ef )
- ( de + ef = df )
- substitution property of equality
- ( ab + bc = de + ef )
- transitive property
- ( ab + bc = de + ef )
- definition of congruence
- substitution property
- reflexive property
- symmetric property
- transitive property
Step1: Recall Segment Addition
We know from the segment addition postulate that for segment \( AC \), \( AB + BC = AC \), and for segment \( DF \), \( DE + EF = DF \).
Step2: Use Given Equalities
We are given \( AC = DF \) and \( BC = DE \). Substitute \( AC = AB + BC \) and \( DF = DE + EF \) into \( AC = DF \), we get \( AB + BC = DE + EF \).
Step3: Substitute \( BC = DE \)
Since \( BC = DE \), we can replace \( BC \) with \( DE \) in the equation \( AB + BC = DE + EF \), resulting in \( AB + DE = DE + EF \).
Step4: Subtract \( DE \) from Both Sides
Using the subtraction property of equality, subtract \( DE \) from both sides of \( AB + DE = DE + EF \). This gives \( AB = EF \).
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\( AB = EF \) (by using segment addition postulate, substitution with given equalities, and subtraction property of equality)