Question

statement reason 1. \\(\overline{de}\\) given 2. if two parallel lines are cut by a transversal, the corresponding angles are congruent. 3. \\(\triangle abc \sim \triangle dbe\\) aa criterion for similarity 4. \\(\frac{ad}{db} = \frac{ce}{eb}\\) corresponding sides of similar triangles are proportional. 5. \\(ab = ad + db\\) \\(cb = ce + eb\\) segment addition 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 10 what is the missing step in this proof? a. \\(\angle cab \cong \angle acb, \angle edb \cong \angle deb\\) b. \\(\angle ade \cong \angle dbe, \angle ced \cong \angle ebd\\) c. \\(\angle cad \cong \angle ace, \angle ade \cong \angle ced\\) d. \\(\angle cab \cong \angle edb, \angle acb \cong \angle deb\\)

Explanation:

The proof starts with $\overline{DE}$ (implied to be parallel to $\overline{AC}$, as it uses parallel lines cut by a transversal for congruent corresponding angles). Step 3 uses the AA similarity criterion, so Step 2 must state the congruent corresponding angles from the parallel lines. For $\triangle ABC \sim \triangle DBE$, the corresponding angles from parallel lines $\overline{DE} \parallel \overline{AC}$ cut by transversals $\overline{AB}$ and $\overline{CB}$ are $\angle CAB \cong \angle EDB$ and $\angle ACB \cong \angle DEB$, which satisfy the AA criterion.

Answer:

D. $\angle CAB \cong \angle EDB, \angle ACB \cong \angle DEB$