Question

proof of basic proportionality theorm​

Answer 1

Answer:

Basic Proptionality Theorm

(B. P. T) :

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

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.

Hope it's helpful for you !