# Rectangular problem

I was trying to solve this problem:

Let P be a point in the interior of rectangle ABCD. Given PA = 3, PD = 4 and PC = 5, find PB.

I feel lost because it's not right to assume P is in the center of the rectangle, I understand it can be anywhere. So what can I learn from lines that connect a vertex to P? The triangles must be similar because they share sides, but I can't prove it. I tried to learn something about the angles of the triangles by drawing parallel lines to the sides of the rectangle, so I could use a lot of properties, but that didn't help me much...

Once I could write four different equations with four variables, I even solved it... but it was the wrong answer. Oh, yeah, I have the answer just as a number, the book says nothing about how to get to it.

The problem seems very simple, but I can't solve it.

[Edit: solved it]

I've drawn this image.

And I've written these equations: $$3^2=a^2+c^2\\4^2=b^2+c^2\\5^2=b^2+d^2\\x^2=a^2+d^2$$ Isolating $c^2$ and $d^2$, I got: $$c^2 = 3^2 - a^2\quad\land\quad c^2 = 4^2 - b^2\\then\\ [I]\quad3^2 - a^2 = 4^2 - b^2\\AND\\ d^2 = 5^2 - b^2\quad\land\quad d^2 = x^2 - a^2\\then\\ [II]\quad5^2 - b^2 = x^2 - a^2$$ I isolated $a^2$ and $b^2$ in $I$ to get a number, which gave me $b^2 - a^2 = 4^2 - 3^2$. And finally I substituted $I$ into $II$, like this: $$5^2 - x^2 = b^2 - a^2\quad\Rightarrow\quad 5^2 - x^2 = 4^2 - 3^2\\x=3\sqrt2$$

But I'm wondering if there's a better way. I didn't like having to write down 4 different equations.

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Hint: Draw lines parallel to $AB$ and $BC$ through $P$ and use Pythagorean theorem to see that $AP^2+CP^2=BP^2+DP^2$
Please, check my solution. I used the parallel lines you told me. I would still like to see the proof to $AP^2 + CP^2 = BP^2 + DP^2$ because I couldn't find that. –  BeetleTheNeato Oct 4 '12 at 17:51
$PA^2 + PC^2 = PB^2 + PD^2$