O is Any Point Inside a Rectangle Abcd. Prove That: Ob2 + Od2 = Oc2 + Oa2. - Mathematics

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Sum

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

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Solution

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Using Pythagorean theorem we have from the above diagram:
OA= AH+ OH= AH+ AE2
OC= CG+ OG= EB+ HD2
OB= EO+ BE= AH+ BE2
OD= HD+ OH= HD+ AE2

Adding these equalities we get:
OA+ OC= AH+ HD+ AE+ EB2
OB+ OD= AH+ HD+ AE+ EB2

From which we prove that for any point within the rectangle there is the relation
OA+ OC= OB+ OD2
Hence Proved.

  Is there an error in this question or solution?
Chapter 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [Page 164]

APPEARS IN

Selina Concise Mathematics Class 9 ICSE
Chapter 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 8 | Page 164

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