Question

Q. solve the following theorem If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.​

Answer 1

Answer:

Given:

In ∆ABC line l || Side BC line l intersects side AB and side AC in P and Q respectively.

To prove :

AP / PB = AQ / QC

Construction:

Draw seg PC and seg QB.

Proof :

A(∆APQ) / A(∆PQB) = AP / PB ----- (I) (Areas are in proportion to the bases)

A(∆APQ) / A(∆PQB) = AQ / QC ------ (II) (Areas are in proportion to the bases)

∆ PQB and ∆ PQC have the same base PQ and PQ || BC, their height is also same.

A(∆ PQB) = A(∆ PQC) ------ (III)

∴ A(∆APQ) / A(∆PQB) = A(∆APQ) / A(∆PQC) ------ from ((I), (II) and (III)

∴ AP / PB = AQ / QC ------- from (I) , (II)

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Answer 2

$\huge\ \: \red{Answer}$

So in the above solution we proved the theorem of Basic Proportionality Theorem Or Thales Theorem.

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