Question

Given: DE and DF are midsegments of ABC

Prove: DE = 1/2 AC

1. midsegments DE and DF: given

2. DF ll BC and AC ll DE: Midsegments are || to the non-included side of the Δ.

3. AB is a transversal cutting DF and BC.

AB is a transversal cutting AC and DE: definition of transversal line

4. m∠FAD = m∠EDB
m∠ADF = m∠DBE: Corresponding Angles Theorem

5. AD = BD: definition of midpoint

6. DBE ≅ ADF: ASA

7. A F = 1/2 AC: definition of midpoint

8.

9. DE = 1/2 AC: Transitive Property of Equality

What is the missing step in this proof?

Answer 1

D. DE = A . F

Reason: Corresponding sides of congruent triangles are congruent.

Explanation:

I got it correct (they won't let me put an A and an F next to each other ???)