The line segment joining the midpoints of any two sides of a triangle s parallel to the third side and is equal to half of it.
Statements 
Reasons 
In △AEF and △CDF 

1. AF=CF 
1.F is midpoint of AC (given) 
2. ∠AFE =∠CFD 
2. Vert.opp.∠s 
3. ∠EAF= ∠DCF 
3. Alt. ∠s,BA∥CD, by the construction and AC is a transversal. 
4. △AEF≌△CDF  4. ASA rule of congruency. 
5. EF=FD and AE=CD 
5. c.p.c.t. 
6. AE=BE 
6. E is the midpoint of AB $($given$)$ 
7. BE=CD 
7. From 5 and 6 
8. EBCD is a parallelogram. 
8. BA∥CD $($const.$)$ and BE=CD $($from step 7$)$ 
9. EF∥BC and ED=BC 
9. Since EBCD is a parallelogram. 
10. EF=½ ED 
10. Since EF=FD, from 5 
11. EF=½BC 
11. Since ED=BC, from 9 
Hence, EF∥BC and EF= ½ BC. 