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
$$\therefore \angle MON=90^o\quad ...(1)$$
In $$\triangle BCD$$
$$E$$ and $$F$$ are midpoints of $$CD$$ and $$BC$$
$$\Rightarrow EF \parallel DB$$ and $$EF=1/2 DB ...(2)\quad$$ [ 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.
$$\therefore \angle EFG=90^o$$