B C
A
D E
SOLUTION
Step 1: ExtendDEtoFso thatDE=EFand joinF C
B C
A
D
E
F
1
2
Step 2: ProveBCF Dis a parallelogram
In△EADand△ECF:
E^ 1 =E^ 2 (vert opp\s=)
AE =CE (given)
DE =EF (by construction)
)△EAD △ECF (SAS)
)ADE^ =CF E^
But these are alternate interior angles, thereforeBD∥F C
BD =DA (given)
DA =F C (△EAD△ECF)
)BD =F C
)BCF Dis a parallelogram (one pair opp. sides= and∥)
ThereforeDE∥BC.
We conclude that the line joining the two mid-points of two sides of a triangle is parallel to the third side.
Step 3: Use properties of parallelogramBCF Dto prove thatDE=^12 BC
DF =BC (opp sides∥m)
andDF = 2 (DE) (by construction)
) 2 DE =BC
)DE =^12 BC
We conclude that the line joining the mid-point of two sides of a triangle is equal to half the length of the third
side.
Chapter 7. Euclidean geometry 263