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Question

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


Solution
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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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