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D, E, F are the midpoints of the sides B...

D, E, F are the midpoints of the sides BC, CA and AB respectively of `triangle`ABC. Prove that
(i) BDEF is a ||gm,
(ii) `ar(triangleDEF)=(1)/(4) ar(triangleABC) and`
(iii) `ar("||gm BDEF") = (1)/(2)ar(triangleABC)`

Text Solution

Verified by Experts

By midpoint theorem, DE||BA and EF||BC.
`therefore "DE||BF and FE||BD"`.
`therefore BDEF " is a ||gm`.
Similarly, AFDE is a ||gm.
And, DCEF is ||gm.
Diagonal DF of ||gm BDEF divides it into two `triangle` of equal area.
`therefore ar(triangleDEF)=ar(triangleBDF).`
Similarly, `ar(triangleDEF)=ar(triangleDCE)`.
And, `ar(triangleDEF)=ar(triangleAFE)`.
`therefore ar(triangleDEF)=ar(BDF)=ar(triangleDCE)=ar(triangleAFE)`
`therefore ar(triangleDEF)=(1)/(4)ar(triangleABC)`.
`=2xx(1)/(4)xxar(triangleABC)=(1)/(2)xxar(triangleABC)`.
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