Question

select the correct answer from each drop - down menu. given: \\(\triangle abc\\) with \\(\overline{de} \parallel \overline{ac}\\) prove \\(\frac{ad}{db} = \frac{cb}{eb}\\) \\(\

$$\begin{array}{|c|c|}\\hline\\text{statements}&\\text{reasons}\\\\ \\hline 1. \\overline{de} \\parallel \\overline{ac}&\\text{given}\\\\ \\hline 2. \\angle cab \\cong \\angle edb, \\angle acb \\cong \\angle deb&\\text{if parallel lines are cut by a transversal, the corresponding angles are congruent.}\\\\ \\hline 3. \\triangle abc \\sim \\triangle dbe&\\\\ \\hline 4. \\frac{ab}{db} = \\frac{cb}{eb}&\\text{corresponding sides of similar triangles are proportional}\\\\ \\hline 5. ab = ad + db, cb = ce + eb&\\\\ \\hline 6. \\frac{ad + db}{db} = \\frac{ce + eb}{eb}&\\text{substitution property of equality}\\\\ \\hline 7. \\frac{ad}{db}+1 = \\frac{ce}{eb}+1&\\text{division}\\\\ \\hline 8. \\frac{ad}{db} = \\frac{ce}{eb}&\\text{subtraction property of equality}\\\\ \\hline\\end{array}$$

\\)

Explanation:

Step1: Analyze Statement 3

To determine the reason for \(\triangle ABC \sim \triangle DBE\), we use the AA (Angle - Angle) similarity criterion. We already have \(\angle CAB\cong\angle EDB\) and \(\angle ACB\cong\angle DEB\) (from the corresponding angles theorem as \(DE\parallel AC\)). So, by AA similarity, \(\triangle ABC \sim \triangle DBE\). The reason for statement 3 is "AA (Angle - Angle) Similarity Criterion".

Step2: Analyze Statement 5

We know that \(AB = AD+DB\) (segment addition postulate, which states that if a point \(D\) lies on a line segment \(AB\), then \(AB=AD + DB\)) and \(CB=CE + EB\) (similarly, by segment addition postulate, since \(E\) lies on \(CB\)). The reason for statement 5 is "Segment Addition Postulate".

Step3: Analyze the Proportion

From the similarity of \(\triangle ABC\) and \(\triangle DBE\) (statement 3), we know that the corresponding sides of similar triangles are proportional. So, \(\frac{AB}{DB}=\frac{CB}{EB}\) (statement 4, reason: corresponding sides of similar triangles are proportional). Then, using the segment addition postulate results (\(AB = AD + DB\) and \(CB=CE + EB\)) and the properties of equality (substitution, division), we can derive \(\frac{AD}{DB}=\frac{CE}{EB}\).

Answer:

For statement 3, the reason is "AA (Angle - Angle) Similarity Criterion"; for statement 5, the reason is "Segment Addition Postulate"; and the final proportion \(\frac{AD}{DB}=\frac{CE}{EB}\) is proven using the properties of similar triangles and segment addition.