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
Join BE, CD.
Draw EF ⊥ AB and DG ⊥ CA
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
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.
Area (ΔDBE) = Area (ΔDCE) -----(3)
Step 4 :
From (1), (2) and (3), we can obtain
AD / DB = AE / EC
Hence, the theorem is proved.
Apart from the proof explained above, if you would like to have practice questions on Basic Proportionality Theorem,
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