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Prove That, If a Line Parallel to a Side of a Triangle Intersect the Other Sides in Two District Points, Then the Line Divides Those Sides in Proportion. - Geometry


Prove that, if a line parallel to a side of a triangle intersects the other sides in two district points, then the line divides those sides in proportion.

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Given: A Δ ABC in which DE||BC, and intersect AB in D and AC in E.

Prove: `"AD"/"BD" = "AE"/"EC"`

Construction: Join BE, CD and draw EF BA and DGCA.


Area ∆ADE = `1/2 xx ("base" xx "height") =1/2("AD. EF")`


Area ∆DBE = `1/2 xx("base" xx "height") =1/2("DB.EF")`


⇒  `("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=[((1/2"AD. EF"))/(1/2"DB. EF"]] = "AD"/"DB"` 


`("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=[((1/2"AE. DG"))/(1/2"EC. DG"]] = "AE"/"EC"` 


ΔDBE and ΔDEC are on the same base DE and b\w the same parallel DE and BC.


 Area (Δ DBE) = Area (ΔDEC)

⇒ `1/("Area"(triangle"DBE")) =1/("Area"(triangle"DEC"))`  .....[Taking reciprocal of both sides]


⇒ `("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=("Area"(triangle"ADE"))/("Area"(triangle"DEC")`  ......[Multiplying both sides by Area (Δ ADE)]

⇒ `"AD"/"DB"="AE"/"EC"`

Hence Proved.

Concept: Property of three parallel lines and their transversals
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