Question

In the adjoining figure the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. prove that AB square + CD square is equal to 80 square + BC square​

Answer 1

$\huge{\textbf{\textsf{{☆ An}}{\purple{sw}}{\pink{er ☆} \: {{}{:}}}}}$

$In △AOB,$

$AB2=OA2+OB2$  

$In △COD,$

$CD2=OC2+OD2$     

$In △OAD,$

$DA2=OA2+OD2$  

$In △BOC,$

$BC2=OC2+OB2$     

$Adding 1 & 2, we get,$

$AB2+CD2=OA2+OB2+OC2+OD2$   

$Adding 3 & 4 , we get,$

$DA2+BC2=OA2+OB2+OC2+OD2$   

$Form 5 & 6, we get,$

$AB2+CD2=BC2+DA2$

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