Question

1. complete the proof below by giving the missing statements and reasons.

given: ac ≅ df, bc ≅ de
prove: ab ≅ ef

Explanation:

Step1: Analyze Given Information

We know \( AC = DF \) and \( BC = DE \). We need to prove \( AB = EF \). Let's recall the segment addition postulate: if a point \( B \) is between \( A \) and \( C \), then \( AC = AB + BC \); similarly, if \( E \) is between \( D \) and \( F \), then \( DF = DE + EF \).

Step2: Apply Segment Addition Postulate

From segment addition:
\( AC = AB + BC \) (since \( B \) is on \( AC \))
\( DF = DE + EF \) (since \( E \) is on \( DF \))

Step3: Substitute Given Equalities

We are given \( AC = DF \) and \( BC = DE \). Substitute \( AC \) with \( AB + BC \) and \( DF \) with \( DE + EF \) in \( AC = DF \):
\( AB + BC = DE + EF \)

Step4: Substitute \( BC = DE \)

Since \( BC = DE \), we can replace \( BC \) with \( DE \) (or vice - versa) in the equation \( AB + BC = DE + EF \). So:
\( AB + DE = DE + EF \)

Step5: Subtract \( DE \) from Both Sides

Subtract \( DE \) from both sides of the equation \( AB + DE = DE + EF \). By the subtraction property of equality, if \( a + b = b + c \), then \( a = c \) (here \( a = AB \), \( b = DE \), \( c = EF \)). So:
\( AB = EF \)

Answer:

By using the segment addition postulate and the properties of equality (substitution and subtraction), we have proven that \( AB = EF \). The final answer is that \( AB \) is equal to \( EF \) (i.e., \( AB = EF \)).