Question

reasoning
introduction to proofs: justifying statements
given: ( ac = df )
( bc = de )
prove: ( ab = ef )
1 ( ac = ab + bc )
2 ( df = de + ef )
3 ( ac = df )
4 ( ab + bc = de + ef )
5 ( bc = de )
6 ( ab + de = de + ef )
7 ( ab = ef )

Explanation:

Step1: Segment addition postulate

By the segment addition postulate, if \(B\) is between \(A\) and \(C\), then \(AC = AB+BC\). Similarly, if \(E\) is between \(D\) and \(F\), then \(DF=DE + EF\).

Step2: Given

The problem states \(AC = DF\) as a given condition.

Step3: Substitution property

Since \(AC = AB + BC\) (from step1), \(DF=DE + EF\) (from step1), and \(AC = DF\) (from step2), we substitute \(AC\) with \(AB + BC\) and \(DF\) with \(DE + EF\) to get \(AB + BC=DE + EF\).

Step4: Given

The problem states \(BC = DE\) as a given condition.

Step5: Substitution property

Substitute \(BC\) with \(DE\) in the equation \(AB + BC=DE + EF\) (from step3) to get \(AB+DE = DE + EF\).

Step6: Subtraction property of equality

Subtract \(DE\) from both sides of the equation \(AB + DE=DE + EF\). Using the subtraction property of equality (\(a + b=b + c\Rightarrow a=c\) when we subtract \(b\) from both sides), we get \(AB=EF\).

Answer:

1. Segment addition postulate
2. Segment addition postulate
3. Given
4. Substitution property
5. Given
6. Substitution property
7. Subtraction property of equality