In the given figure, $ABCD$ is a kite in which $BC=CD,AB=AD$ and $E,F,G$ are midpoints of $CD,BC$ and $AB$ respectively. Prove that, $\angle EFG={90}^{\circ }$

It is given that

$ABCD$ is a kite in which $BC=CD,AB=AD$

$E,F,G$ are midpoints of $CD,BC$ and $AB$

To prove :

$\angle EFG=90^o$

Construction :

Join $AC$ and $BD$

Construct $GH$ through $G$ paralled to $FE$

Proof:

We know that,

Diagonals of a kite intersect at right angles

$\therefore \angle MON=90^o...(1)$

In $△BCD$

$E$ and $F$ are midpoints of $CD$ and $BC$

$\Rightarrow EF\parallel DB$ and $EF=1/2DB...(2)$ [ By basic proportionality theorem ]

Now, $EF\parallel DB\Rightarrow MF\parallel ON$

Similarly, $FG\parallel CA\Rightarrow FN\parallel MO$

Therefore, in quadrilateral $MFNO$,

$MF\parallel ON,FN\parallel MO$  and  $\angle MON=90^o$

$\Rightarrow MFNO$ is a square.

$\therefore \angle EFG=90^o$