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.