Since you have a rectangle, the length is half the width. So $\displaystyle BC$ and $\displaystyle AD$ are 8.

And you can make this picture:

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Now apply the Pythagorean Theorem a bunch of times.

$\displaystyle (XB)^2 + (BC)^2 = (XC)^2$

$\displaystyle 8^2 + 10^2 = (XC)^2$

$\displaystyle 164 = (XC)^2$

$\displaystyle XC = 2\sqrt{41}$

Then apply it again: $\displaystyle (XA)^2 + (AD)^2 = (DX)^2$

$\displaystyle 6^2 + 8^2 = (DX)^2$

$\displaystyle 36 + 64 = (DX)^2$

$\displaystyle 80 = (DX)^2$

$\displaystyle DX = 4\sqrt{5}$

Now let's fill that in our pic:

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Since $\displaystyle XZ$ is an angle bisector, then $\displaystyle ZC = DZ$

So both are 8.

Apply Pythagorean Theorem to get $\displaystyle XZ$

$\displaystyle (ZC)^2 + (XZ)^2 = (2\sqrt{41})^2$

$\displaystyle 64 + (XZ)^2 = 164$

$\displaystyle (XY)^2 = 100$

$\displaystyle XY = 10$

And without more info, I don't see how you can find $\displaystyle XY$