Question
For the following statments state whether true $(T)$ or false$(F):$If $O$ is any point inside a rectangle $ABCD$ then $OA^2 + OC^2 = OB^2 + OD^2$
✓
Answer
True.
Solution:
Construction: Draw $EF || AB$ through $O.$
In $\triangle\text{OEA}$ and $\triangle\text{OFC},$ by pythagoras theorem,
$OA^2 = OE^2 + AE^2$ and $OC^2= OF^2 + CF^2$
Adding the two equations, we get
$OA^2 + OC^2 = OE^2 + AE^2 + OF^2 + CF^2 .....(i)$
$\triangle \text{OFB}$ and $\triangle \text{ODE},$ by pythagoras theorem,
$OB^2 = OF^2 + FB^2 $ and $OD^2 = OE^2 + DE^2$
Adding the two equation, we get
$OB^2 + OD^2 = OF^2 + FB^2 + OE^2 + DE^2 .....(ii)$
By Construction since $EF || CD,$
$DE = CF$ and $AE = FB$
So, from (i) and (ii), we have
$OA^2 + OC^2 = OB^2 + OD^2$
Solution:
Construction: Draw $EF || AB$ through $O.$
In $\triangle\text{OEA}$ and $\triangle\text{OFC},$ by pythagoras theorem,
$OA^2 = OE^2 + AE^2$ and $OC^2= OF^2 + CF^2$
Adding the two equations, we get
$OA^2 + OC^2 = OE^2 + AE^2 + OF^2 + CF^2 .....(i)$
$\triangle \text{OFB}$ and $\triangle \text{ODE},$ by pythagoras theorem,
$OB^2 = OF^2 + FB^2 $ and $OD^2 = OE^2 + DE^2$
Adding the two equation, we get
$OB^2 + OD^2 = OF^2 + FB^2 + OE^2 + DE^2 .....(ii)$
By Construction since $EF || CD,$
$DE = CF$ and $AE = FB$
So, from (i) and (ii), we have
$OA^2 + OC^2 = OB^2 + OD^2$
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