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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Solution

Through O, draw EOF || AB. Then, ABFE is a rectangle.

In right triangles OEA and OFC, we have:

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

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

∴ ${O A}^{2}$ + ${O C}^{2}$ = ${O E}^{2}$ + ${O F}^{2}$ + ${A E}^{2}$ + ${C F}^{2}$ ---------(i)

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

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

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

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

= ${O E}^{2}$ + ${O F}^{2}$ + ${A E}^{2}$ + ${C F}^{2}$ --------------(ii) [∴ BF = AE & DE = CF]

From (i) and (ii), we get the result.

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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.

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