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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.

$$\therefore \angle EFG=90^o$$

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