Question

O is any point inside a rectangle ABCD. Prove that $O{B}^2+O{D}^2=O{A}^2+O{C}^2$.

Answer 1

Through O, draw PQ || BC so that P lies on AB and Q lies on DC.

Now, PQ||BC.

Therefore, PQ ⊥ AB and PQ ⊥ DC (∠ B = ${90}^{\circ }$ and ∠ C = ${90}^{\circ }$)

So, ∠ BPQ = ${90}^{\circ }$ and ∠ CQP = ${90}^{\circ }$

Therefore, BPQC and APQD are both rectangles.

Now, from $△$ OPB, ${OB}^2$ = ${BP}^2$ + ${OP}^2$ (1)

Similarly, from $△$ OQD,

${OD}^2$ = ${OQ}^2$ + ${DQ}^2$ (2)

From $△$ OQC, we have

${OC}^2$ = ${OQ}^2$ + ${CQ}^2$ (3)

and from $△$OAP, we have

${OA}^2$ = ${AP}^2$ + ${OP}^2$ (4)

Adding (1) and (2),

${OB}^2$ + ${OD}^2$ = ${BP}^2$ + ${OP}^2$ + ${OQ}^2$ + ${DQ}^2$

= ${CQ}^2$ + ${OP}^2$ + ${OQ}^2$ + ${AP}^2$

(As BP = CQ and DQ = AP)

= ${CQ}^2$ + ${OQ}^2$ + ${OP}^2$ + ${AP}^2$

= ${OC}^2$ + ${OA}^2$ [From (3) and (4)]