O is any point inside a rectangle ABCD. Prove that `OB^(2)+OD^(2)=OA^(2)+OC^(2)`.
DEDUCTION In the given figure, O is a point inside a rectangle ABCD such that `OB=6cm, OD=8 cm and OA=5 cm,` find the length of OC.

Explanation

GIVEN O is point inside a rectangle ABCD.
TO PROVE `OB^(2)+OD^(2)=OA^(2)+OC^(2)`.
CONSTRUCTION Through O, drawn `POQ||BC` so that P lies on AB and Q lies on DC.

PROOF We have
`POQ||BC rArr PQ bot AB and QP bot DC`
`rArr angle BPQ =90^(@) and angle CQP=90^(@)`
`:. BP=CQ` [ oppositie sides of a rectangle]
`DQ=AP` [ opposite sides of a rectangle]
From right `Delta OPB`, we have `OD^(2)=OP^(2)+BP^(2)`
From right `Delta OQD`, we have `OD^(2)=OQ^(2)+DQ^(2)`
From right `Delta OPA`, we have `OA^(2)=OP^(2)+AP^(2)`
From right `Delta OQC`, we have `OC^(2)=OQ^(2)+CQ^(2)`.
`:. OB^(2)+OD^(2)=OP^(2)+OQ^(2)+BP^(2)+DQ^(2)= OP^(2)+OQ^(2)+CQ^(2)+AP^(2) [ :. BP =CQ and DQ= AP]`
`=(OP^(2)+AP^(2))+(OQ^(2)+CQ^(2))`
`=OA^(2)+OC^(2)`
Hence, `OB^(2)+OD^(2)=OA^(2)+OC^(2)`
DEFUCTION Let OC=xm
`OB^(2)+OD^(2)=OA^(2)+OC^(2)`
`rArr 6^(2)+8^(2)=5^(2)+x^(2)`
`rArr x^(2)=36+64-25=75`
`rArr x=sqrt(75)=5sqrt(3)=(5xx1.732)=8.66 rArr OC =8.66` cm