Question

Verified by Toppr

A triangle of $ABC$ and $D,E,F$ are the mid-points of sides $BC,CA$ and $AB$ respectively.

To prove :

$△AFE≅△FBD≅△EDC≅△DEF$.

Proof :

Since the segment joining the mid-points of the sides of a triangle is half of the third side. Therefore,

$DE=21 AB⟹DE=AF=BF$ ..........(1)

$EF=21 BC⟹EF=BD=CD$ .........(2)

$DF=21 AC⟹DF=AE=EC$ ..........(3)

Now, in $△s$ $DEF$ and $AFE,$

$DE=AF$

$DF=AE$

and, $EF=FE$

So, by SSS criterion of congruence,

$△DEF≅△AFE$

Similarly, $△DEF≅△FBD$ and $△DEF≅△EDC$

Hence, $△AFE≅△FBD≅△EDC≅△DEF$

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