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Questions
Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
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.
Solution
Given: Let a ΔABC in which a line DE parallel to BC intersects AB at D and AC at E.
To Prove: DE divides the two sides in the same ratio.
`"AD"/"DB" = "AE"/"EC"`
Construction: Join BE, CD and draw EF ⊥ AB and DG ⊥ AC.
Proof: Here,
Area of triangle = `1/2` × base × height
Area of ΔADE = `1/2` × AD × EF
or
Area of ΔADE = `1/2` × AE × DG
Similarly,
Area of ΔBDE = `1/2` × DB × EF
Area of ΔDEC = `1/2` × EC × DG
`"ar(ΔADE)"/"ar(ΔBDE)" = (1/2 × "AD" × "EF")/(1/2 × "DB" × "EF")`
`"ar(ΔADE)"/"ar(ΔBDE)" = "AD"/"DB"` ...(1)
From (2) and (4),
`"ar(ΔADE)"/"ar(ΔDEC)" = (1/2 × "AE" × "DG")/(1/2 × "EC" × "DG")`
`"ar(ΔADE)"/"ar(ΔDEC)" = "AE"/"EC"` ...(2)
Since, ΔBDE and ΔDEC lie between the same parallel DE and BC and on the same base DE.
∴ ar(ΔBDE) = ar(ΔDEC) ...(3)
From (1), (2) and (3), we get,
`"AD"/"BD" ="AE"/"EC"`
Hence proved.
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