Question

‘O’ is any point inside a rectangle ABCD. Prove that OB² + OD² = OA² + OC²​

Answer 1

Answer:

HEY BUDDY HERE IS YOUR ANSWER:-

Step-by-step explanation:

Given, ABCD is a rectangle. O is any point inside the rectangle.

Construction: Construct perpendiculars OP, OQ, OR, OS on each side. 

Now, OA2+OC2=(AS2+OS2)+(OQ2+QC2) (Considering right triangles ASO and COQ)

But AS=BQ and QC=SD, Therefore,

OA2+OC2=(BQ2+OS2)+(OQ2+SD2)

OA2+OC2=(BQ2+OQ2)+(OS2+SD2)

OA2+OC2=OB2+OD2 (Considering right triangles OSD and OBQ)

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Answer 2

Answer:

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° and ∠C = 90°) So, ∠BPQ = 90° and ∠CQP = 90° Therefore, BPQC and APQD are both rectangles. Now, from ΔOPB, OB2 = BP2 + OP2   ....