Basic Proportionality Theorem | Thales Theorem | BPT

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Proof of BPT | Proof of Basic Proportionality Theorem | Proof of BPT
CBSE Grade 10 Mathematics, Proof of Triangles Theorem
Basic Proportionality Theorem
Thales Theorem
If a line is drawn parallel to one side of a triangle to intersect the other two sides in two distinct points, then the other two sides are divided in the same ratio.
Given: In ∆ ABC, DE ∥ BC
To prove: AD/DB = AE/EC
Construction: Draw EN ⊥ AD, DM ⊥ AE, join BE and CD.
Proof: Area of ∆ ADE = 1/2 × Base × Height
Area of ∆ ADE = 1/2 × AD × EN
Also Area of ∆ BDE = 1/2 × DB × EN
Dividing, (ar (ADE))/(ar (BDE)) = (1/2 × AD × EN)/(1/2 × DB × EN) = AD/DB …………(1)
Again, Area of ∆ ADE = 1/2 × AE × DM
Also Area of ∆ CDE = 1/2 × EC × DM
Dividing, (ar (ADE))/(ar (CDE)) = (1/2 × AE × DM)/(1/2 × EC × DM) = AE/EC …………(2)
Now ∆ BDE and ∆ CDE are on the same base DE and between the same parallels
DE and BC.
∴ ar (BDE) = ar (CDE)
So from (1) and (2), AD/DB = AE/EC
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