Question

given: $be \parallel cd$ prove: $\frac{bc}{ab} = \frac{ed}{ae}$
triangle proof

statements	reasons
2. $\angle a \cong \angle a$	2. reflexive property
3. $\angle acd \cong \angle abe$	3. corresponding angles formed by parallel lines and a transversal are $\cong$.
4. $\angle adc \cong \angle aeb$	4. corresponding angles formed by parallel lines and a transversal are $\cong$.
5. $\triangle abe \sim \triangle acd$	5. aa similarity postulate
6. $\frac{ac}{ab} = \frac{ad}{ae}$	6. definition of similar triangles
7. $ac = ab + bc$, $ad = ae + ed$	7. segment addition postulate
8. $\frac{ab + bc}{ab} = \frac{ae + ed}{ae}$	8. substitution
9. $\frac{ab}{ab} + \frac{bc}{ab} = \frac{ae}{ae} + \frac{ed}{ae}$	9. addition
10. $\frac{bc}{ab} = \frac{ed}{ae}$	10. ???

fill in the missing reason for the proof.

a transitive property
b subtraction property
c sss similarity theorem
d segment addition postulate

Explanation:

To find the missing reason for step 10, we analyze the previous steps. In step 9, we have \(\frac{AB}{AB}+\frac{BC}{AB}=\frac{AE}{AE}+\frac{ED}{AE}\). We know that \(\frac{AB}{AB} = 1\) and \(\frac{AE}{AE}=1\). If we subtract 1 from both sides of the equation \(\frac{AB}{AB}+\frac{BC}{AB}=\frac{AE}{AE}+\frac{ED}{AE}\), we get \(\frac{BC}{AB}=\frac{ED}{AE}\). This operation is based on the Subtraction Property of Equality, which states that if \(a + b=a + c\), then \(b = c\) (by subtracting \(a\) from both sides).

Now let's analyze the other options:

• Option A: The Transitive Property states that if \(a = b\) and \(b = c\), then \(a = c\). This is not relevant here as we are not using transitivity.
• Option C: The SSS Similarity Theorem is about the ratio of sides in similar triangles, but we are not using it here.
• Option D: The Segment Addition Postulate is about adding segments to form a longer segment, which is not related to this step.

So the correct reason is the Subtraction Property.

Answer:

B. Subtraction Property