**Basic Proportionality Theorem Proof : **

In this section, we are going to see, how to prove Basic Proportionality Theorem.

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.

Apart from the proof explained above, if you would like to have practice questions on Basic Proportionality Theorem,

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