A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that: ${O A}^{2} + {O C}^{2} = {O B}^{2} + {O D}^{2}$ . [4 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark
Proof : 2 Marks

Through O, draw $E F ∥ A B$ . Then, ABFE is a rectangle.

In right triangles OEA and OFC, we have:

${O A}^{2} = {O E}^{2} + {A E}^{2}$

${O C}^{2} = {O F}^{2} + {C F}^{2}$

$∴ {O A}^{2} + {O C}^{2} = {O E}^{2} + {O F}^{2} + {A E}^{2} + {C F}^{2} \dots \dots (1)$

Again, in right triangles OFB and OED, we have:

${O B}^{2} = {O F}^{2} + {B F}^{2}$

${O D}^{2} = {O E}^{2} + {D E}^{2}$

$∴ {O B}^{2} + {O D}^{2} = {O F}^{2} + {O E}^{2} + {B F}^{2} + {D E}^{2}$

$\Rightarrow {O B}^{2} + {O D}^{2} = {O E}^{2} + {O F}^{2} + {A E}^{2} + {C F}^{2} \dots \dots (2)$

[ $∵$ BF = AE & DE = CF]

From (1) and (2), we get

${O A}^{2} + {O C}^{2} = {O B}^{2} + {O D}^{2}$ .

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A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that: ${O A}^{2}$ + ${O C}^{2}$ = ${O B}^{2}$ + ${O D}^{2}$ .

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A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that: OA2+OC2=OB2+OD2. [4 MARKS]

A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that: ${O A}^{2} + {O C}^{2} = {O B}^{2} + {O D}^{2}$ . [4 MARKS]