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Updated on : 2022-09-05

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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 :

$∠EFG=90_{o}$

Construction :

Join $AC$ and $BD$

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

Proof:

We know that,

Diagonals of a kite intersect at right angles

$∴∠MON=90_{o}...(1)$

In $△BCD$

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

$⇒EF∥DB$ and $EF=1/2DB...(2)$ [ By basic proportionality theorem ]

Now, $EF∥DB⇒MF∥ON$

Similarly, $FG∥CA⇒FN∥MO$

Therefore, in quadrilateral $MFNO$,

$MF∥ON,FN∥MO$ and $∠MON=90_{o}$

$⇒MFNO$ is a square.

$∴∠EFG=90_{o}$

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