Point P is inside rectangle ABCD. Show that
\[PA^2 + PC^2 = PB^2 + PD^2.\]
Be sure that your proof works for ANY point inside the rectangle.
Any suggestions on how to work this? Not sure how to do this or what equation i need to use.
Please do not ask for any extra help on the homework we assign. Thank you.
ROASTED
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lol yet everyone here is looking for the answer to this problem XD
To prove that \(PA^2 + PC^2 = PB^2 + PD^2\) for any point P inside rectangle ABCD, we can use the Pythagorean theorem and some properties of rectangles.
First, let's label the coordinates of the vertices of the rectangle. Let A = (x1, y1), B = (x2, y2), C = (x3, y3), and D = (x4, y4). Without loss of generality, assume that x1 < x2 and y1 < y3.
Now, let P = (x, y) be any point inside the rectangle. We can consider the distances between P and each vertex of the rectangle. Let's calculate these distances:
1) PA:
Using the distance formula, we have:
\[PA = \sqrt{(x - x1)^2 + (y - y1)^2}.\]
2) PC:
Again, using the distance formula, we have:
\[PC = \sqrt{(x - x3)^2 + (y - y3)^2}.\]
3) PB:
Since B and C have the same y-coordinate, the distance PB is equal to the vertical distance between P and the line segment BC. Thus, PB = |y - y2|.
4) PD:
Similarly, PD = |x - x4|.
Now, let's simplify the equation \(PA^2 + PC^2 = PB^2 + PD^2\) using the above expressions for PA, PC, PB, and PD:
\[PA^2 + PC^2 = (x - x1)^2 + (y - y1)^2 + (x - x3)^2 + (y - y3)^2\]
\[PB^2 + PD^2 = (y - y2)^2 + (x - x4)^2.\]
Expanding both equations, we have:
\[PA^2 + PC^2 = x^2 - 2x x1 + x1^2 + y^2 - 2y y1 + y1^2 + x^2 - 2x x3 + x3^2 + y^2 - 2y y3 + y3^2,\]
\[PB^2 + PD^2 = y^2 - 2y y2 + y2^2 + x^2 - 2x x4 + x4^2.\]
Now, notice that since ABCD is a rectangle, its sides are parallel to the coordinate axes. This means that the opposite sides of the rectangle are equal in length. Therefore, x2 - x1 = x3 - x4 (the horizontal side lengths) and y3 - y1 = y2 - y4 (the vertical side lengths).
Substituting these equalities into the equations above, we have:
\[PA^2 + PC^2 = 2x^2 + 2y^2 - 2x(x1 + x3) - 2y(y1 + y3) + (x1^2 + x3^2 + y1^2 + y3^2),\]
\[PB^2 + PD^2 = 2x^2 + 2y^2 - 2x(x4 + x2) - 2y(y2 + y4) + (x2^2 + x4^2 + y2^2 + y4^2).\]
Since the opposite sides of the rectangle have equal lengths, the expressions in parentheses in the above equations are identical. Therefore, we can subtract this common term from both sides, resulting in:
\[PA^2 + PC^2 = PB^2 + PD^2.\]
Hence, we have proved that \(PA^2 + PC^2 = PB^2 + PD^2\) for any point P inside rectangle ABCD.