Point P lies in the interior of rectangle ABC | vedclass

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(N/A) Let the rectangle be $ABCD$ in the Cartesian plane. Let $A = (0, b)$,$B = (a, b)$,$C = (a, 0)$,and $D = (0, 0)$.Let the coordinates of point $P$

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(N/A) Let the rectangle be $ABCD$ in the Cartesian plane. Let $A = (0, b)$,$B = (a, b)$,$C = (a, 0)$,and $D = (0, 0)$.
Let the coordinates of point $P$ be $(x, y)$.
Using the distance formula,the squared distances are:
$PA^2 = (x - 0)^2 + (y - b)^2 = x^2 + y^2 - 2by + b^2$
$PC^2 = (x - a)^2 + (y - 0)^2 = x^2 - 2ax + a^2 + y^2$
$PB^2 = (x - a)^2 + (y - b)^2 = x^2 - 2ax + a^2 + y^2 - 2by + b^2$
$PD^2 = (x - 0)^2 + (y - 0)^2 = x^2 + y^2$
Now,calculate the sums:
$PA^2 + PC^2 = (x^2 + y^2 - 2by + b^2) + (x^2 - 2ax + a^2 + y^2) = 2x^2 + 2y^2 - 2ax - 2by + a^2 + b^2$
$PB^2 + PD^2 = (x^2 - 2ax + a^2 + y^2 - 2by + b^2) + (x^2 + y^2) = 2x^2 + 2y^2 - 2ax - 2by + a^2 + b^2$
Since both sums are equal,$PA^2 + PC^2 = PB^2 + PD^2$ is proved.

Point $P$ lies in the interior of rectangle $ABCD$. Prove that $PA^2 + PC^2 = PB^2 + PD^2$.

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