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?

Draw —DE , which is a midsegment of △ABC. 0. 1. 2. 3. 4. 5. 6. −1. −2. 0. 1. 2. A. C ... 3. How are the midsegments of a triangle related to the sides of the triangle? 4. ... D(1.5, 4.5). E(5, 3). Segments. BC = 4. AC = 7.62. AB = 7.07. DE = ? DF = ? ... Given —. DE is a midsegment of △OBC. Prove —. DE —. OC and DE = 1— 2.

Step-by-step explanation:

im alright what do you need

step-by-step explanation:

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads