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# 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, $$\angle EFG={ 90 }^{ \circ }$$

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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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