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In the adjoining figure D, E and F are t...

In the adjoining figure D, E and F are the mid-points of the sides BC, CA and AB respectively of `Delta ABC`. Prove that:
(i) `square BDEF` is a parallelogram
(ii) area of `Delta DEF = (1)/(4) xx " area of " Delta ABC`
(iii) `square BDEF = (1)/(2) xx " area of " Delta ABC`

Text Solution

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(i) In `Delta ABC`
Since, F is the mid-point of AB and E is the mid-point of AC
`{:( :." "FE ||BC),(rArr" "FE ||BD),("Similarly "ED||FB):}` (mid - point theorem)
`:. Square BDEF` is a parallelogram (`:.` pairs of opposite sides are parallel)
(ii) Since, `square BDEF` is a parallelogram and FD is its diagonal
`:.` area of `Delta FBD = " area of " Delta DEF`..(1)
Similarly, we can prove that
area of `Delta CDE = " area of " Delta DEF`...(2)
and area of `Delta AFE = " area of " Delta DEF`...(3)
`:. 4 xx " area of " Delta DEF = " area of "Delta DEF + " area of " Delta DEF + " area of "Delta DEF + " area of "Delta DEF`
`= " area of " Delta DEF + " area of "Delta AEF + " area of "Delta FBD + " area of "Delta CDE`
= area of `Delta ABC`
`rArr " area of " Delta DEF = (1)/(4) xx " area of " Delta ABC`
(iii) Now, area of `square BDEF = 2 xx " area of " Delta ABC`
(`:.` diagonal divides the parallelogram into two equal areas)
`= 2 xx (1)/(4) xx " area of " Delta ABC`
`= (1)/(2) xx " area of "Delta ABC`
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