Question

given: \\( \frac { d c } { e c } = \frac { a d } { b e } \\) prove: \\( \overline { a b } \parallel \overline { d e } \\) complete the steps of the proof. 1. \\( \frac { d c } { e c } = \frac { a d } { b e } \\) 1. given 2. \\( \frac { d c } { e c } + 1 = \frac { a d } { b e } + 1 \\) 2. addition property 3. \\( \frac { d c } { e c } + \frac { d c } { d c } = \frac { a d } { b e } + \frac { e c } { e c } \\) 3. property of proportion 4. \\( \frac { a d + d c } { d c } = \frac { b e + e c } { e c } \\) 4. addition of fractions 5. \\( a c = a d + d c \\) 5. segment addition 6. \\( b c = b e + e c \\) 6. postulate 7. \\( \frac { a c } { d c } = \frac { b c } { e c } \\) 7. substitution property 8. \\( \triangle a b c \sim \triangle d e c \\) 8. \\( \triangle \\) 9. \\( \angle 1 = \angle 2 \\) 9. \\( \angle \\)

Explanation:

Step1: Analyze step 2

From step 1 $\frac{DC}{EC}=\frac{AD}{BE}$, add 1 to both sides.
$$\frac{DC}{EC}+1=\frac{AD}{BE}+1$$

Step2: Analyze step 3

Use the property of proportion $\frac{a}{b}+\frac{c}{c}=\frac{a + c}{b}$ (here $a = AD,b = BE,c=BE$ for the right - hand side and $a = DC,b = EC,c = EC$ for the left - hand side)
$$\frac{DC}{EC}+\frac{EC}{EC}=\frac{AD}{BE}+\frac{BE}{BE}$$

Step3: Analyze step 4

By the addition of fractions $\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}$ (reverse), and segment addition ($AC=AD + DC$ and $BC=BE + EC$)
$$\frac{AD+DC}{DC}=\frac{BE + EC}{EC}$$

Step4: Analyze step 5

By segment addition postulate ($AC=AD + DC$ and $BC=BE + EC$)
$$AC=AD + DC,BC=BE + EC$$

Step5: Analyze step 6

By substitution property (substitute $AC$ for $AD + DC$ and $BC$ for $BE + EC$ in step 4)
$$\frac{AC}{DC}=\frac{BC}{EC}$$

Step6: Analyze step 7

By reflexive property ($\angle3=\angle3$)

Step7: Analyze step 8

By the Side - Angle - Side (SAS) similarity criterion (from step 6 $\frac{AC}{DC}=\frac{BC}{EC}$ and step 7 $\angle3=\angle3$)
$$\triangle ABC\sim\triangle DEC$$

Step8: Analyze step 9

By the corresponding angles of similar triangles (since $\triangle ABC\sim\triangle DEC$, $\angle1$ and $\angle2$ are corresponding angles)
$$\angle1=\angle2$$

Answer:

$\angle1=\angle2$