Question

in a rectangle ABCD, P is any interior point. then prove that PA^2+PC^2=PB^2+PD^2.​

Answer 1

Given : a rectangle ABCD, and P is any interior point.

To prove : $PA^2+PC^2=PB^2+PD^2$

Step-by-step explanation:

Let the coordinate of,

A = ( 0, 0 ), B = (u, 0), C=(u, v), D=(0, v) and P=(x, y).

Then,

$PA^2+PC^2=(x^2+y^2)+\{(x-u)^2+(y-v)^2\}\hfill (1)$

$PB^2+PD^2=\{(x-u)^2+y^2\}+\{x^2+(y-v)^2\}\hfll (2)$

On compairing (1) and (2) we get,

$PB^2+PD^2=\{(x-u)^2+y^2\}+\{x^2+(y-v)^2\}=(x^2+y^2)+\{(x-u)^2+(y-v)^2\}=PA^2+PC^2$

Hence proved.

Answer 2

Explanational mention below

Step-by-step explanation:

Given : ABCD is a rectangle, and P is any interior point.

To prove : $PA^2+PC^2=PB^2+PD^2$

Proof: Let the coordinate of,

A = ( 0, 0 ), B = (u, 0), C=(u, v), D=(0, v) and P=(x, y).

Then,

$PA^2+PC^2=(x^2+y^2)+\{(x-u)^2+(y-v)^2\}\hfill (1)$

$PA^2+PC^2= x^2+y^2+ x^2 -2xu +u^2 +y^2 -2vy +v^2 \\= 2x^2 + 2y^2+v^2+u^2 -2xu-2vy.......... (ii)$

$PB^2+PD^2=\{(x-u)^2+y^2\}+\{x^2+(y-v)^2\}\hfll ..............(2)$

$PB^2+PD^2= x^2 -2xu +u^2 +y^2 + x^2 +y^2-2yv+v^2 \\ =2x^2 +2Y^2 + u^2+v^2-2xu-2yv.......... (ii)$

On comparing (i) and (i) we get,

$PA^2+PC^2=PB^2+PD^2$

To learn more

i)If abcd be a rectangle and p be any point in the plane of the rectangle,then prove that pa^2+pc^2=pb^2+pd^2

https://brainly.in/question/2021243

ii)O is any point inside a rectangle ABCD. prove that OB^2 + OD^2 = OA^2 + OC^2

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