Question

State and prove "Basic Proportionality Theorem."

Approach

State and prove the Basic Proportionality Theorem (Thales' theorem). Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Given: In $△ A BC$ , $D E ∥ BC$ . To prove: $\frac{A D}{D B} = \frac{A E}{EC}$ . Construction: Draw perpendiculars $D M ⊥ A C$ and $EN ⊥ A B$ ; join $BE$ and $C D$ . Proof: area $(△ A D E) = \frac{1}{2} A D \cdot EN$ and area $(△ D BE) = \frac{1}{2} D B \cdot EN$ , so $\frac{ar ( △ A D E )}{ar ( △ D BE )} = \frac{A D}{D B}$ . Similarly, $\frac{ar ( △ A D E )}{ar ( △ D CE )} = \frac{A E}{EC}$ . Since $D E ∥ BC$ , triangles $D BE$ and $D CE$ have equal areas (same base $D E$ , same parallels). Hence the ratios are equal.

Step-by-step working

1
Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
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Given: In $△ A BC$ , $D E ∥ BC$ . To prove: $\frac{A D}{D B} = \frac{A E}{EC}$ . Construction: Draw $D M ⊥ A C$ , $EN ⊥ A B$ ; join $BE$ , $C D$ . [Figure shown.]
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Proof: $\frac{ar ( △ A D E )}{ar ( △ D BE )} = \frac{\frac{1}{2} A D \cdot EN}{\frac{1}{2} D B \cdot EN} = \frac{A D}{D B}$ .
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Similarly, $\frac{ar ( △ A D E )}{ar ( △ D CE )} = \frac{\frac{1}{2} A E \cdot D M}{\frac{1}{2} EC \cdot D M} = \frac{A E}{EC}$ .
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Since $D E ∥ BC$ , triangles $D BE$ and $D CE$ lie on the same base $D E$ and between the same parallels, so their areas are equal. Hence $\frac{A D}{D B} = \frac{A E}{EC}$ .
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Concepts used

basic proportionality theoremprovethales theorem

Approach

Step-by-step working

Concepts used

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