BASIC PROPORTIONALITY THEOREM PROOF

If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then  it divides the two sides in the same ratio. 

Given : In a triangle ABC, a straight line l parallel to BC, intersects AB at D and AC at E.  

To prove : AD/DB  =  AE/EC

Construction :

Join BE, CD. 

Draw EF ⊥ AB and DG ⊥ CA

Proof :

Step 1 :

Because EF ⊥ AB, EF is the height of the triangles ADE and DBE. 

Area (ΔADE)  =  1/2 ⋅ base ⋅ height  =  1/2 ⋅ AD ⋅ EF

Area (ΔDBE)  =  1/2 ⋅ base ⋅ height  =  1/2 ⋅ DB ⋅ EF

Therefore, 

Area (ΔADE) / Area (ΔDBE)  :

=  (1/2 ⋅ AD ⋅ EF) / (1/2 ⋅ DB ⋅ EF)

Area (ΔADE) / Area (ΔDBE)  =  AD / DB -----(1)

Step 2 : 

Similarly, we get

Area (ΔADE) / Area (ΔDCE)  :

=  (1/2 ⋅ AE ⋅ DG) / (1/2 ⋅ EC ⋅ DG)

Area (ΔADE) / Area (ΔDCE)  =  AE / EC -----(2)

Step 3 : 

But ΔDBE and ΔDCE are on the same base DE and between the same parallel straight lines BC and DE. 

Therefore, 

Area (ΔDBE)  =  Area (ΔDCE) -----(3)

Step 4 : 

From (1), (2) and (3), we can obtain

AD / DB  =  AE / EC

Hence, the theorem is proved.