Question

given: △abc with de || ac
prove: \\(\frac{ad}{db} = \frac{ce}{eb}\\)

triangle abc with d on ab, e on bc, de || ac

statements	reasons
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\\)
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}\\)	substitution property of equality
7. \\(\frac{ad}{db} + 1 = \frac{ce}{eb} + 1\\)	division
8. \\(\frac{ad}{db} = \frac{ce}{eb}\\)	subtraction property of equality

Explanation:

Step1: Identify Reason for Step 3

To determine the reason for $\triangle ABC \sim \triangle DBE$, we use the AA (Angle - Angle) similarity criterion. From step 2, we know that $\angle CAB\cong\angle EDB$ and $\angle ACB\cong\angle DEB$. If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar by AA similarity. So the reason for step 3 is "AA (Angle - Angle) Similarity Criterion".

Step2: Identify Reason for Step 5

The statements $AB = AD + DB$ and $CB=CE + EB$ are based on the segment addition postulate, which states that if a point $D$ lies on segment $AB$, then $AB=AD + DB$, and similarly for segment $CB$ with point $E$. So the reason for step 5 is "Segment Addition Postulate".

Step3: Verify Step 7

Starting from $\frac{AD + DB}{DB}=\frac{CE + EB}{EB}$, we can split the fractions. Using the property $\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}$, we have $\frac{AD+DB}{DB}=\frac{AD}{DB}+\frac{DB}{DB}=\frac{AD}{DB}+ 1$ and $\frac{CE + EB}{EB}=\frac{CE}{EB}+\frac{EB}{EB}=\frac{CE}{EB}+1$. So the operation here is the division of the numerator terms (or the property of splitting a fraction into two fractions with the same denominator), which is why the reason is "Division (or Fraction Splitting Property)".

Step4: Verify Step 8

Starting from $\frac{AD}{DB}+1=\frac{CE}{EB}+1$, we subtract 1 from both sides of the equation. By the subtraction property of equality, if $a + c=b + c$, then $a = b$. Here, $a=\frac{AD}{DB}$, $b = \frac{CE}{EB}$, and $c = 1$. So subtracting 1 from both sides gives $\frac{AD}{DB}=\frac{CE}{EB}$, which is consistent with the subtraction property of equality.

Answer:

• Step 3 Reason: AA (Angle - Angle) Similarity Criterion
• Step 5 Reason: Segment Addition Postulate
• Step 7 Explanation: Using $\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}$, $\frac{AD + DB}{DB}=\frac{AD}{DB}+\frac{DB}{DB}=\frac{AD}{DB}+1$ and $\frac{CE + EB}{EB}=\frac{CE}{EB}+\frac{EB}{EB}=\frac{CE}{EB}+1$
• Step 8 Verification: Subtract 1 from both sides of $\frac{AD}{DB}+1=\frac{CE}{EB}+1$ (subtraction property of equality) to get $\frac{AD}{DB}=\frac{CE}{EB}$