Question

. D, E, F are the mid-points of sides BC, CA and AB respectively of ∆ABC.Then ∆DEF is congruent to triangle ; *


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Answer 1

$AF = \frac{1}{2} AB \: \\ |since F \: is \: midpoint \: of \: AB\: |$

and,

$DE = \frac{1}{2} AB \\ |by \: midpoint \: theoram|$

Combining these equations we get,

AF = DE ....................(i)

similarly,

AE = DF ....................(ii)

Now, in ∆AFE and ∆DFE,

FE = FE [ Common]

AF = DE [From eqⁿ(i)]

AE = DF [From eqⁿ(ii)]

So, ∆AFE = ∆DFE [SSS]