D and E are the mid-points of the sides AB an | vedclass

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(C) To prove that $CF$ is equal and parallel to $DA$,we consider $	riangle ADE$ and $	riangle CFE$.In these triangles:$1$. $AE = CE$ (Since $E$ is t

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$DE = EF$

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(C) To prove that $CF$ is equal and parallel to $DA$,we consider $	riangle ADE$ and $	riangle CFE$.In these triangles:$1$. $AE = CE$ (Since $E$ is the mid-point of $AC$)$2$. $\angle AED = \angle CEF$ (Vertically opposite angles)If we are given $DE = EF$,then by $SAS$ congruence criterion,$	riangle ADE \cong 	riangle CFE$.By $CPCT$,we get $AD = CF$ and $\angle ADE = \angle CFE$.Since $\angle ADE$ and $\angle CFE$ are alternate interior angles,it implies $AD \parallel CF$.Thus,the additional information required is $DE = EF$.

$D$ and $E$ are the mid-points of the sides $AB$ and $AC$ respectively of $\triangle ABC$. $DE$ is produced to $F$. To prove that $CF$ is equal and parallel to $DA$,we need an additional information which is

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