Question

consider the proof.
given: segment ab is parallel to line de.
prove: \\(\frac{ad}{dc} = \frac{be}{ec}\\)
image of a triangle with points a, b, c, d, e and angles 1, 2, 3, 4

1. \\(ab \parallel de\\) - given
2. \\(\angle 1\\) and \\(\angle 3\\) are corresponding \\(\angle s\\); \\(\angle 2\\) and \\(\angle 4\\) are corresponding \\(\angle s\\) - def of corresponding angles
3. \\(\angle 1 \cong \angle 3, \angle 2 \cong \angle 4\\) - corresponding \\(\angle s\\) post
4. \\(\triangle abc \sim \triangle dec\\) - aa similarity post

5.? -

1. \\(ac = ad + dc\\); \\(bc = be + ec\\) - segment addition postulate
2. \\(\frac{ad + dc}{dc} = \frac{be + ec}{ec}\\) - substitution
3. \\(\frac{ad}{dc} + \frac{dc}{dc} = \frac{be}{ec} + \frac{ec}{ec}\\) - addition of fractions
4. \\(\frac{ad}{dc} + 1 = \frac{be}{ec} + 1\\) - simplification of fractions
5. \\(\frac{ad}{dc} = \frac{be}{ec}\\) - subtraction property

what is the missing statement in step 5?
\\(\bigcirc\\) \\(ac = bc\\)
\\(\bigcirc\\) \\(\frac{ac}{dc} = \frac{bc}{ec}\\)
\\(\bigcirc\\) \\(ad = be\\)
\\(\bigcirc\\) \\(\frac{ad}{dc} = \frac{be}{ec}\\)

Explanation:

Step1: Recall similar triangle properties

Similar triangles have proportional corresponding sides. Step4 states $\triangle ABC \sim \triangle DEC$ (AA similarity).

Step2: Identify proportional sides

For similar triangles $\triangle ABC$ and $\triangle DEC$, the ratio of corresponding sides $\frac{AC}{DC} = \frac{BC}{EC}$ holds. This connects Step4 (similarity) to Step6 (segment addition).

Answer:

B. $\frac{AC}{DC} = \frac{BC}{EC}$