Question

in the following triangle, ed || cb.
below is the proof that \\(\frac{ec}{ea} = \frac{db}{ad}\\). the proof is divided into two parts, where the title of each part indicates its main purpose.
complete part b of the proof.
part a: prove \\(\frac{ac}{ae} = \frac{ab}{ad}\\)
show the steps.
part b: prove \\(\frac{ec}{ea} = \frac{db}{ad}\\)
statement reason
6 \\(ac = ae + ec\\) segment addition postulate
7 \\(ab = ad + db\\) segment addition postulate
8 \\(\frac{ae + ec}{ae} = \\) pick expression substitution (part a, 6, 7)
9 \\(1 + \frac{ec}{ae} = 1 + \frac{db}{ad}\\) pick reason (9)
10 \\(\frac{ec}{ae} = \frac{db}{ad}\\) subtract 1 from both sides. (10)

Explanation:

Step1: Analyze Step 8

From Part A, we have \(\frac{AC}{AE}=\frac{AB}{AD}\). And from steps 6 and 7, \(AC = AE + EC\) and \(AB=AD + DB\). So substitute \(AC\) and \(AB\) into the proportion from Part A. So \(\frac{AE + EC}{AE}=\frac{AD + DB}{AD}\).

Step2: Analyze Step 9

We can rewrite \(\frac{AE+EC}{AE}\) as \(\frac{AE}{AE}+\frac{EC}{AE}=1 + \frac{EC}{AE}\) and \(\frac{AD + DB}{AD}=\frac{AD}{AD}+\frac{DB}{AD}=1+\frac{DB}{AD}\). So the reason for step 9 is that we split the fractions (using the property \(\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}\)) and since \(\frac{AE}{AE} = 1\) and \(\frac{AD}{AD}=1\), we get \(1+\frac{EC}{AE}=1+\frac{DB}{AD}\). The reason here is the "Addition of fractions" (or splitting the fraction into two parts) and substitution from step 8.

Step3: Analyze Step 10

If we subtract 1 from both sides of the equation \(1+\frac{EC}{AE}=1+\frac{DB}{AD}\), we get \(\frac{EC}{AE}=\frac{DB}{AD}\) (by the subtraction property of equality: if \(a + c=b + c\), then \(a = b\)).

For Step 8: The expression should be \(\frac{AD + DB}{AD}\) (since \(AC=AE + EC\) and \(AB = AD+DB\), and from Part A \(\frac{AC}{AE}=\frac{AB}{AD}\), so substituting gives \(\frac{AE + EC}{AE}=\frac{AD + DB}{AD}\)).

For Step 9: The reason is "Splitting the fraction (or using \(\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}\)) and substitution" (more precisely, the property of fractions: \(\frac{x + y}{x}=\frac{x}{x}+\frac{y}{x}=1+\frac{y}{x}\) for \(x
eq0\)).

Answer:

• Step 8: \(\boldsymbol{\frac{AD + DB}{AD}}\)
• Step 9: Reason is "Property of fractions ( \(\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}\) ) and substitution" (or "Splitting the fraction into two terms")
• Step 10: Follows from subtracting 1 from both sides (subtraction property of equality) to get \(\frac{EC}{EA}=\frac{DB}{AD}\)