Question

O is any point inside a rectangle ABCD as shown in the figure. Prove that OB2 + OD2 = OA2 + OC2.

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Answer 1

$\huge\underline\mathrm{Question}:-$

O is any point inside a rectangle ABCD as shown in the figure. Prove that OB² + OD² = OA² + OC²

$\huge\underline\mathrm{Solution}:-$

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

PQ || BC

Therefore, PQ ⊥ AB and PQ ⊥ DC (∠ B = 90° and ∠ C = 90°)

So, ∠ BPQ = 90° and ∠ CQP = 90°

Hence, BPQC and APQD are both rectangles.

By Pythagoras theorem,

In ∆ OPB,

OB2 = BP² + OP²…..(1)

Similarly,

In ∆ OQD,

OD² = OQ² + DQ²…..(2)

In ∆ OQC,

OC² = OQ² + CQ²…..(3)

In ∆ OAP,

OA² = AP¹ + OP²…..(4)

Adding (1) and (2),

OB²+ OD² = BP² + OP² + OQ² + DQ²

= CQ² + OP² + OQ² + AP²

(since BP = CQ and DQ = AP)

= CQ² + OQ² + OP² + AP²

= OC² + OA² [From (3) and (4)]

Hence proved that OB² + OD² = OA² + OC²