# 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 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)

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.

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