anwar ali has a farm in the shape of a square abcd . its diagonals ac and bd intersect at o

show that triangle aob = boc=cod=doa

Answer :

We draw Anwar Ali's square shaped farm ABCD , As :

We know sides of square equal in length , So
AB  =  BC   =  CD  =  DA 
And

We know " The diagonals of a square are the same length , And they are bisect each other ."

So ,

OA  =  OB  =  OC  =  OD 

Now In $∆$ AOB  , $∆$ BOC , $∆$ COD And $∆$ DOA

AB  =  BC  =  CD  =  DA  ( Sides of square )

OA  =  OB  =  OC  =  OD  ( As we know diagonals bisect each other )

So,

$∆$ AOB $\cong$ $∆$ BOC $\cong$ $∆$ COD $\cong$$∆$ DOA  ( By SSS rule )

We can say that

These triangle , AS :

$∆$ AOB  = $∆$ BOC = $∆$ COD = $∆$ DOA  ( Hence proved )