# Theorem Of Parallelograms

## Mid-Point Theorem

The line segment joining the mid-points of any two sides of a triangle s parallel to the third side and is equal to half of it.

Mid-point theorem
The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.
Given. A triangle ABC,E and F are the mid -points of sides AB and AC respectively.
To prove: EF ∥BC and EF=½ BC.
Construction: Through C, draw a line parallel to BA to meet EF produced at D.

Proof:
 Statements Reasons In △AEF and △CDF 1. AF=CF 1.F is mid-point 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 mid-point 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.