Prove that :
“If a line parallel to a side of a triangle intersects the remaining sides in two distince points, then the line divides the sides in the same proportion.”
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Solution
Given :In ΔABC 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) Δ PQB and Δ PQC have the same base PQ and PQ || BC,
their height is also same.
∴ A(Δ PQB) = A(Δ PQC) ..... (III)
∴` (A(APQ))/(A(PQB)) =( A(APQ))/(A(PQC))` ........ from ((I), (II) and (III)
∴ `(AP)/(PB) = (AQ)/(QC)` ........ from (I) , (II)
Concept: General Equation of a Line
Is there an error in this question or solution?
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