Question

statements reasons 1. \\(\overline{de} \parallel \overline{ac}\\) given 2. \\(\angle cab \cong \angle edb\\) \\(\angle acb \cong \angle deb\\) if parallel lines are cut by a transversal, the corresponding angles are congruent. 3. \\(\triangle abc \sim \triangle dbe\\) aa criterion for similarity 4. \\(\frac{ab}{db} = \frac{cb}{eb}\\) corresponding sides of similar triangles are proportional 5. \\(ab = ad + db\\) \\(cb = ce + eb\\) 6. \\(\frac{ad + db}{db} = \frac{ce + eb}{eb}\\) 7. \\(\frac{ad}{db} + 1 = \frac{ce}{eb} + 1\\) 8. \\(\frac{ad}{db} = \frac{ce}{eb}\\) addition property of equality segment addition sum of sides of similar triangles subtraction property of equality

Explanation:

Step1: Analyze Statement 5

Statement 5 shows \( AB = AD + DB \) and \( CB = CE + EB \). This is the application of the segment addition postulate, which states that if a point lies on a segment, the sum of the two smaller segments formed is equal to the length of the original segment. So the reason for statement 5 is the segment addition postulate.

Step2: Analyze Statement 6

From statement 4, we have \( \frac{AB}{DB}=\frac{CB}{EB} \). In statement 5, we found \( AB = AD + DB \) and \( CB = CE + EB \). Substituting these into the proportion from statement 4 gives \( \frac{AD + DB}{DB}=\frac{CE + EB}{EB} \). This is a substitution of equal quantities (since \( AB = AD + DB \) and \( CB = CE + EB \)) into the proportion, so the reason for statement 6 is substitution (substituting \( AB \) with \( AD + DB \) and \( CB \) with \( CE + EB \) in the proportion \( \frac{AB}{DB}=\frac{CB}{EB} \)).

Step3: Analyze Statement 7

We start with \( \frac{AD + DB}{DB}=\frac{CE + EB}{EB} \). We can split the fractions: \( \frac{AD}{DB}+\frac{DB}{DB}=\frac{CE}{EB}+\frac{EB}{EB} \). Since \( \frac{DB}{DB} = 1 \) and \( \frac{EB}{EB}=1 \), this simplifies to \( \frac{AD}{DB}+ 1=\frac{CE}{EB}+1 \). This is the result of splitting the numerator of each fraction (using the property \( \frac{a + b}{c}=\frac{a}{c}+\frac{b}{c} \)) and simplifying \( \frac{DB}{DB} \) and \( \frac{EB}{EB} \) to 1.

Step4: Analyze Statement 8

Starting from \( \frac{AD}{DB}+ 1=\frac{CE}{EB}+1 \), we can subtract 1 from both sides of the equation (subtraction property of equality). Subtracting 1 from both sides gives \( \frac{AD}{DB}=\frac{CE}{EB} \). So the reason for statement 8 is the subtraction property of equality (if \( a + c=b + c \), then \( a = b \) when we subtract \( c \) from both sides).

Answer:

• Reason for statement 5: Segment addition postulate (or segment addition)
• Reason for statement 6: Substitution (substituting \( AB = AD + DB \) and \( CB = CE + EB \) into \( \frac{AB}{DB}=\frac{CB}{EB} \))
• Reason for statement 7: Splitting the fraction (using \( \frac{a + b}{c}=\frac{a}{c}+\frac{b}{c} \)) and simplifying \( \frac{DB}{DB}=1 \), \( \frac{EB}{EB}=1 \)
• Reason for statement 8: Subtraction property of equality (subtracting 1 from both sides of \( \frac{AD}{DB}+ 1=\frac{CE}{EB}+1 \))