State and prove basic proportionality theorem.
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Answers
→ Theorem
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 :- ΔABC, BC ∥ DE, DE intersects sides AB and AC.
To prove that :-
Construction :- Join BE and CD, draw DM ⊥ AC
Proof :- We know that are of triangle = 1/2 x b x h
- Area of ΔADE = 1/2 x AD x EN
similarly,
- Area of ΔBDE = 1/2 x DB x EN
and,
- Area of ΔDEC = 1/2 x EC x DM
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Therefore,
Similarly,
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ΔBDE and ΔDEC are on the same bases and between the same parallel DE, Therefore their area will be equal.
- arΔBDE = arΔDEC
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From (1) and (2)
↬ Proved.
★ If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio or are proportional. ★
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Given : In ∆ABC, PQ || BC
To prove : AP / PB = AQ / QC
Construction : Draw QH perpendicular to AB, Join BQ and PC. [ You can take one more perpendicular ].
Proof : Area of ∆APQ = 1/2 × base × height
= 1/2 × AP × QH
Similarly, Area ∆PBQ = 1/2 × base × height
= 1/2 × PB × QH
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Ratio of their areas :
Area ∆APQ / Area ∆PBQ = AP / PB ( 1/2 , 1/2 and QH, QH are cancelled out ) _____(1)
Similarly,
Area ∆APQ / Area ∆PQC = AQ / QC _____(2)
Now,
∆PBQ and ∆PQC are on the same base and lying between same parallel lines,
Therefore,
Area ∆PBQ = Area ∆PQC _____(3)
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From (1), (2) and (3) :
We can write : AP / PB = AQ / QC
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Therefore,
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio or are proportional.