maths

$D$ is the midpoint of $AB$

So, $AD$ =$DB$= $CF$= $2c $

Similarly $AE$ =$EC$=$2b $

Let $∠AED=α$ $ADE=β$and$DAE=γ$

Since $AD$ is parallel to $FC$

$∠FEC=α∠EFC=β$and$∠ECF=γ$

Applying $S.A.S$ Congruency for triangle$ADE$ and $CFE$, $AD=CF$, $AE=AC$

$∠DAE=∠ECF$

So $Δ_{k}DAE$ and$Δ_{k}FCE$ are congruent

So $DE=EF$

It is given that $DE=EF$,then by $S.S.S.$ congruency all angles will be equal and $CF$ and parallel to $DA$

$∴$ Additionsl information needed is $DE=EF$.

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