Question

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

Explanation:

Step1: Analyze the given proportion

We have \(\frac{AD}{DC}=\frac{BE}{EC}\) (given).

Step2: Use the addition property of equality

Add \(1\) to both sides: \(\frac{AD}{DC}+ 1=\frac{BE}{EC}+1\).

Step3: Simplify the left - hand side

\(\frac{AD}{DC}+1=\frac{AD + DC}{DC}\) (by the rule \(\frac{a}{b}+1=\frac{a + b}{b}\)).

Step4: Simplify the right - hand side

\(\frac{BE}{EC}+1=\frac{BE + EC}{EC}\) (by the rule \(\frac{a}{b}+1=\frac{a + b}{b}\)).

Step5: Use segment addition

Since \(AC=AD + DC\) and \(BC=BE + EC\) (segment addition postulate).

Step6: Substitute

We get \(\frac{AC}{DC}=\frac{BC}{EC}\) (substitution property).

Step7: Analyze angles

\(\angle3=\angle3\) (reflexive property).

Step8: Prove similarity

By the SAS (Side - Angle - Side) similarity theorem, \(\triangle ABC\sim\triangle DEC\) (because \(\frac{AC}{DC}=\frac{BC}{EC}\) and \(\angle3\) is common).

Step9: Conclude parallel lines

If \(\triangle ABC\sim\triangle DEC\), then \(\angle1=\angle2\) (corresponding angles of similar triangles). And if \(\angle1=\angle2\), then \(\overline{AB}\parallel\overline{DE}\) (alternate interior angles are equal).

Answer:

The proof uses the SAS similarity theorem. First, start with the given \(\frac{AD}{DC}=\frac{BE}{EC}\), use the addition property of equality, segment addition postulate, substitution property, reflexive property of angles, and then apply the SAS similarity theorem to show \(\triangle ABC\sim\triangle DEC\). Finally, use the property of similar triangles (corresponding angles) to conclude \(\overline{AB}\parallel\overline{DE}\).