Given: ΔABC with AD:DB = CE:EB Prove DE||AC

1. AD:DB = CE:EB Given

2. AD:DB+1 = CE:EB+1 Addition Property of Equality

3. (AD+DB)/DB = (CE+EB)/EB Using common denominators

4. AB = AD+ DB and CB = CE +EB segment addition

5. AB:DB = CB:EB Substitution property of equality

6. [Fill in missing statement from choices below] Reflexive property of congruence

7. ΔABC ~ ΔDBE SAS similarity criterion

8. ∠BAC ≅ ∠BDE Corresponding angles of similar triangles are congruent.

9. DE||AC If the corresponding angles formed by two lines cut by a transversal are congruent, then the lines are parallel.

What is the missing step in this proof?

A. ∠ABC ≅ ∠DBE

B. ∠BCA ≅ ∠BDE

C. ∠ACB ≅ ∠DEB

D. ∠BDE ≅ ∠ADE

E. ∠CAB ≅ ∠DAC

I got this wrong on the test and I don't know what I did wrong.

In order to invoke the SAS similarity criterion, the angle between the proportional sides must be congruent in the two triangles. Here, that angle is the same angle (angle B) in both triangles, which is why the reflexive property is cited. It tells you ∠B ≅ ∠B.

The only answer choice involving angle B is the first one:

A. ∠ABC ≅ ∠DBE

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