Let's first draw a diagram to better understand the problem:

C
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4 | \
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|-------\
P B
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A

In this diagram, P is the interior point of the right-angled isosceles triangle ABC, and the perpendicular distances from P to AB, BC, and CA are all 4.

Let's denote the length of each side of the triangle as x. Since the triangle is right-angled and isosceles, we have:

AC = BC = x
AB = √(x^2 + x^2) = √2x

Since P is an interior point, we can draw perpendiculars from P to each side of the triangle:

C
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| \
| \
4 | \
| \
| \
| \
|-------\
P B
| /|
| / |
| / |
| / |
| / |
| / |
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|/ |
A D

Let D be the foot of the perpendicular from P to AB.

Since D is the foot of the perpendicular, we can say that AD = 4 and DB = 4.

Now, let's consider the right-angled triangle APD. Using the Pythagorean theorem, we have:

AP^2 = AD^2 + PD^2
AP^2 = 4^2 + 4^2
AP^2 = 16 + 16
AP^2 = 32
AP = √32 = 4√2

Similarly, considering the right-angled triangle BPD, we have:

BP^2 = BD^2 + PD^2
BP^2 = 4^2 + 4^2
BP^2 = 16 + 16
BP^2 = 32
BP = √32 = 4√2

Finally, considering the right-angled triangle CPD, we have:

CP^2 = CD^2 + PD^2
CP^2 = (AC - AP)^2 + PD^2
CP^2 = (x - 4√2)^2 + 4^2
CP^2 = x^2 - 8√2x + 32 + 16
CP^2 = x^2 - 8√2x + 48

Since we are given that CP = 4, we can set up the equation:

4^2 = x^2 - 8√2x + 48
16 = x^2 - 8√2x + 48
x^2 - 8√2x + 32 = 0

Now, we can solve this quadratic equation for x. Using the quadratic formula, we have:

x = [8√2 ± √((8√2)^2 - 4(1)(32))] / (2)
x = [8√2 ± √(