Probability stick I drop parallel to diagonal of rectangle fits within the rectangle Here's a question from my probability textbook:

A floor is paved with rectangular bricks each $a$ inches long and $b$ inches wide. A stick $c$ inches long is thrown upon the floor so as to fall parallel to a diagonal of the bricks. Show that the chance that it lies entirely on one brick is $(a^2 + b^2 - c^2) \div (a^2 + b^2)$.

Here's what I did. So let's say we have a rectangle of height $a$ and width $b$. Let's say our stick of length $c$ falls parallel to the diagonal going from top left to bottom right. We want to calculate the probability that it lies entirely within this rectangle. Each stick fall is determined by its top left point. If you draw this all out, the part of our $a$ by $b$ rectangle where if the upper left point of our stick is dropped the stick fits entirely in our rectangle, is the upper left part of the rectangle with height $a - {{ac}\over{\sqrt{a^2 + b^2}}}$ and width $b - {{bc}\over{\sqrt{a^2 + b^2}}}$. So the probability we want to calculate is$${{\left(a - {{ac}\over{\sqrt{a^2 + b^2}}}\right)\left(b - {{bc}\over{\sqrt{a^2 + b^2}}}\right)}\over{ab}} = \left(1 - {c\over{\sqrt{a^2 + b^2}}}\right)^2 = 1 - {{2c}\over{\sqrt{a^2 + b^2}}} + {{c^2}\over{a^2 + b^2}}.$$However, it's not clear how or if this even equals ${{a^2 + b^2 - c^2}\over{a^2 + b^2}}$. Can anyone help me show that they are equal, or if they aren't, let me know where specifically I went wrong?
EDIT: I'm still not convinced by Math Lover's answer. Let's use the diagram he's using. Let's draw rectangle $FCC'F'$, where $F'$ lies on $AC$, $C'$ lies on $CB$, $FF'$ is perpendicular to $DC$, $F'C'$ is perpendicular to $CB$. I claim our desired probability is the area of $FCC'F'$ divided by the area of $ABCD$.
For every stick of length $c$ that falls that's parallel to $AC$, its upper-right endpoint determines it. And only if I select a point in rectangle $FCC'F'$ will its extension by length $c$ in the bottom-left direction parallel to $AC$ lie fully within rectangle $ABCD$.
The length of $FC$ is the length of $DC$ minus the length of $DF$, so following the convention Math Lover uses of the length of $DC$ being $a$ and the length of $DA$ being $b$, the length of $FC$ is $a -{{ac}\over{\sqrt{a^2 + b^2}}}$. Similarly, the length of $CC'$ is $b - {{bc}\over{\sqrt{a^2 + b^2}}}$. Multiplying those two and dividing by $ab$ gets the probability I calculated before the edit, which differs from what Math Lover got.
However, this still differs from the answer we're supposed to find, so can anyone elaborate in further detail Math Lover's initial two sentences in his answer?

The problem is in how you are calculating the area of the region where the stick should fall. You are multiplying side lengths but the region is not a rectangle.

 A: 
The midpoint of the stick should fall inside the red rectangle. If it falls into the blue area, then the stick will touch the borders.
$$a' = c a/(2\sqrt{a^2+b^2})$$
$$b' = c b/(2\sqrt{a^2+b^2})$$
$$
\text{Probability} = \frac{\text{Red area}}{\text{Area of large blue rectangle}} = \frac{(a-2 a')(b-2 b')}{a\, b}
$$
Using the results for $a'$ and $b'$:
$$
\text{Probability} = \frac{\left(\sqrt{a^2+b^2}-c\right)^2}{a^2+b^2} = \frac{c^2}{a^2+b^2}-\frac{2 c}{\sqrt{a^2+b^2}}+1
$$
Therefore, I think that the answer reported in the question (namely, $(a^2 + b^2 - c^2) \div (a^2 + b^2)$) is wrong, unless this is some kind of Bertrand paradox. Just to be clear:
$$
\frac{a^2 + b^2 - c^2}{ a^2 + b^2} \neq
\frac{\left(\sqrt{a^2+b^2}-c\right)^2}{a^2+b^2}
$$
unless $c=0$ or $c = \sqrt{a^2+b^2}$: in those extreme cases the probability is 1 and 0, respectively.
NOTE: if the stick is thrown on the diagonal (and not on a generic line parallel to the diagonal), then using the same argument of the midpoint, it's easy to obtain
$$
\text{Probability} =\frac{\text{diagonal}-(c/2)-(c/2)}{\text{diagonal}}= \frac{ \sqrt{a^2+b^2}-c }{\sqrt{a^2+b^2}} 
$$
As @Radost wrote in the comments, it is interesting to note that this probability is exactly the square root of the one we obtained before (it's just an elegant scaling argument: the area of similar rectangles scales with the square of the diagonal).
A: Your calculations seem to be correct; you end up with this messy expression:
${{\left(a - {{ac}\over{\sqrt{a^2 + b^2}}}\right)\left(b - {{bc}\over{\sqrt{a^2 + b^2}}}\right)}\over{ab}} = \left(1 - {c\over{\sqrt{a^2 + b^2}}}\right)^2 = 1 - {{2c}\over{\sqrt{a^2 + b^2}}} + {{c^2}\over{a^2 + b^2}}$
(Sorry I'm not good at mathjax...) Multiply each term so the denominator of all is $a^2+b^2$, and combine terms, getting the numerator:
$(a^2+b^2) - (2c\sqrt{a^2+b^2}) + (c^2)$ ...(1)
Then notice that if you write $x$ for $\sqrt{a^2+b^2}$, you get
$x^2 - 2cx + c^2$
Corrected...
which is obviously $(x - c)^2$, and not $x^2 - c^2$ as the original answer would imply. Thus the original answer is wrong.
Apologies for the wrong answer!
