Question

by substitution, \\(\frac{ba + cb}{cb} = \frac{de + cd}{cd}\\).
by aa similarity, \\(\triangle ace \sim \triangle bcd\\).
simplifying, \\(\frac{ba}{cb} + 1 = \frac{de}{cd} + 1\\).
from the definition of similar polygons, \\(\frac{ca}{cb} = \frac{ce}{cd}\\). by the segment addition postulate, \\(ca = ba + cb\\) and \\(ce = de + cd\\)
we are given \\(\overline{bd} \parallel \overline{ae}\\), so \\(\angle 4 \cong \angle 1\\) and \\(\angle 3 \cong \angle 2\\) because corresponding angles are congruent.
thus, \\(\frac{ba}{cb} = \frac{de}{cd}\\) by subtracting 1 from each side.
rewriting each side as a sum, \\(\frac{ba}{cb} + \frac{cb}{cb} = \frac{de}{cd} + \frac{cd}{cd}\\).

Explanation:

Step1: Identify the starting point

The problem involves proving triangle similarity and related proportionality. The first step in a proof about similar triangles with parallel lines should establish the congruent angles from the parallel lines. So the first step is: "We are given \(\overline{BD} \parallel \overline{AE}\), so \(\angle 4 \cong \angle 1\) and \(\angle 3 \cong \angle 2\) because corresponding angles are congruent."

Step2: Apply AA Similarity

After establishing two pairs of congruent angles, we use AA (Angle - Angle) Similarity to state that \(\triangle ACE \sim \triangle BCD\). So the next step is: "By AA Similarity, \(\triangle ACE \sim \triangle BCD\)."

Step3: Use the definition of similar polygons

From the definition of similar polygons, the ratios of corresponding sides are equal. So we have \(\frac{CA}{CB}=\frac{CE}{CD}\). Also, by the Segment Addition Postulate, \(CA = BA + CB\) and \(CE=DE + CD\). So this step is: "From the definition of similar polygons, \(\frac{CA}{CB}=\frac{CE}{CD}\). By the Segment Addition Postulate, \(CA = BA + CB\) and \(CE = DE + CD\)."

Step4: Substitute the segment addition results

Substitute \(CA = BA + CB\) and \(CE = DE + CD\) into the proportion \(\frac{CA}{CB}=\frac{CE}{CD}\). We get \(\frac{BA + CB}{CB}=\frac{DE + CD}{CD}\). So this step is: "By substitution, \(\frac{BA+CB}{CB}=\frac{DE + CD}{CD}\)."

Step5: Rewrite as a sum of fractions

Rewrite each side of the equation \(\frac{BA + CB}{CB}=\frac{DE + CD}{CD}\) as a sum of two fractions. \(\frac{BA}{CB}+\frac{CB}{CB}=\frac{DE}{CD}+\frac{CD}{CD}\), and since \(\frac{CB}{CB} = 1\) and \(\frac{CD}{CD}=1\), we have \(\frac{BA}{CB}+1=\frac{DE}{CD}+1\). So this step is: "Rewriting each side as a sum, \(\frac{BA}{CB}+\frac{CB}{CB}=\frac{DE}{CD}+\frac{CD}{CD}\) (or \(\frac{BA}{CB}+1=\frac{DE}{CD}+1\))."

Step6: Simplify the equation

Subtract 1 from both sides of the equation \(\frac{BA}{CB}+1=\frac{DE}{CD}+1\). We get \(\frac{BA}{CB}=\frac{DE}{CD}\). So this step is: "Thus, \(\frac{BA}{CB}=\frac{DE}{CD}\) by subtracting 1 from each side."

To order the steps correctly for a proof (assuming we are proving \(\frac{BA}{CB}=\frac{DE}{CD}\) given \(\overline{BD}\parallel\overline{AE}\)):

1. We are given \(\overline{BD} \parallel \overline{AE}\), so \(\angle 4 \cong \angle 1\) and \(\angle 3 \cong \angle 2\) because corresponding angles are congruent.
2. By AA Similarity, \(\triangle ACE \sim \triangle BCD\).
3. From the definition of similar polygons, \(\frac{CA}{CB}=\frac{CE}{CD}\). By the Segment Addition Postulate, \(CA = BA + CB\) and \(CE = DE + CD\).
4. By substitution, \(\frac{BA+CB}{CB}=\frac{DE + CD}{CD}\).
5. Rewriting each side as a sum, \(\frac{BA}{CB}+\frac{CB}{CB}=\frac{DE}{CD}+\frac{CD}{CD}\) (or \(\frac{BA}{CB}+1=\frac{DE}{CD}+1\)).
6. Thus, \(\frac{BA}{CB}=\frac{DE}{CD}\) by subtracting 1 from each side.

If we assume the question is to order these steps in the correct logical order for a proof, the correct order is:

1. We are given \(\overline{BD} \parallel \overline{AE}\), so \(\angle 4 \cong \angle 1\) and \(\angle 3 \cong \angle 2\) because corresponding angles are congruent.
2. By AA Similarity, \(\triangle ACE \sim \triangle BCD\).
3. From the definition of similar polygons, \(\frac{CA}{CB}=\frac{CE}{CD}\). By the Segment Addition Postulate, \(CA = BA + CB\) and \(CE = DE + CD\).
4. By substitution, \(\frac{BA+CB}{CB}=\frac{DE + CD}{CD}\).
5. Rewriting each side as a sum, \(\frac{BA}{CB}+1=\frac{DE}{CD}+1\).
6. Thus, \(\frac{BA}{CB}=\frac{DE}{CD}\) by subtracting 1 from each side.

Answer:

The correct order of the steps for the proof is:

1. We are given \(\overline{BD} \parallel \overline{AE}\), so \(\angle 4 \cong \angle 1\) and \(\angle 3 \cong \angle 2\) because corresponding angles are congruent.

1. By AA Similarity, \(\triangle ACE \sim \triangle BCD\).

1. From the definition of similar polygons, \(\frac{CA}{CB}=\frac{CE}{CD}\). By the Segment Addition Postulate, \(CA = BA + CB\) and \(CE = DE + CD\).

1. By substitution, \(\frac{BA+CB}{CB}=\frac{DE + CD}{CD}\).

1. Rewriting each side as a sum, \(\frac{BA}{CB}+1=\frac{DE}{CD}+1\).

1. Thus, \(\frac{BA}{CB}=\frac{DE}{CD}\) by subtracting 1 from each side.