Question

given: ( cf = eg ) ( cd = ef ) prove: ( df = fg ) 4 ( eg = ef + fg ) segment addition property 5 ( cd = ef ) addition and subtraction properties 6 angle addition property

Explanation:

Step1: Analyze the given information

We know \(CF = EG\) and \(CD = EF\). We need to prove \(DF=FG\).

Step2: Use the Segment Addition Postulate

By the Segment Addition Postulate, \(CF=CD + DF\) (since \(C - D - F\)) and \(EG=EF + FG\) (since \(E - F - G\)).

Step3: Substitute the known equal - lengths

Since \(CF = EG\) and \(CD = EF\), we substitute into the equations from Step 2. So \(CD + DF=EF + FG\).

Step4: Apply the Subtraction Property of Equality

Subtract \(CD\) (which is equal to \(EF\)) from both sides of the equation \(CD + DF=EF + FG\). We get \(DF=FG\).

Answer:

To correct the proof:

• For statement 5, the reason should be "Given" (since \(CD = EF\) is given in the problem, not derived from addition and subtraction properties at this stage).
• For statement 6, the correct statement should be \(CF=CD + DF\) (by Segment Addition Property) and then substitute \(CF\) with \(EG\) (because \(CF = EG\)) to get \(EG=CD + DF\), and since \(CD = EF\), we have \(EG=EF + DF\). Then, using \(EG=EF + FG\) (Segment Addition Property), by the Transitive Property \(EF + DF=EF + FG\), and by the Subtraction Property of Equality \(DF = FG\).