Prove that a triangle is right triangle I'd like to know if there's any theorem to prove that the triangle ACB' is a right triangle and that the angle ACB' is 90°.

We know that ACB and A'B'C'  are right triangles, so in my opinion ACB' is also a right triangle, but I don't know how to prove it
If there's any theorem or explanation please let me know.
 A: Assuming that is a right prism, we have that $\;CC'\perp\Delta ABC\;$  and since this last is a right triangle at $\;C\;$ , we then get that $\;AC\perp CBB'C'\;$ , as $\;AC\;$ is perpendicular to two different lines on that plane that go through the line's foot on the plane (point $\;C\;$ ) , namely: to $\;CC'\;$ and to $\;BC\;$ . 
From here it follows that $\;AC\;$ must be perpendicular to any line contained in plane $\;CBB'C'\;$ and thus also to $\;CB'\;$ .
The above is just vector geometry in three dimensions.
A: If we just have have to give a statement (not proving) that $\angle(ACB')=90°$ , You can say that:
Since the planes $ ACC'A'$ and $CBB'C'$ are perpendicular to each other,
$\implies $ Every line segment contained by $ACC'A' $ must also be perpendicular to line contained by $CBB'C$.
This way we interpret $AC$ perpendicular to $CB'$ or $\angle ACB' = 90° $

But as the question is not just giving the figure, but  lengths of sides are also given , it dictates us to use those given data, and prove in a much simpler way just using Pythagoras Theorem and its converse.
For proving $$ \angle ACB' = 90° $$
You will need Converse of Pythagoras Theorem, which states that in any triangle, if the square of one side is equal to the sum of squares of other two sides, then the angle opposite to larger side (hypotenuse) is a right angle

Please note one thing that $\color{green}{CC'B'B}$ is a rectangle. 
(Which seems as a parallelogram in your $3D$ figure)
With $$BC= 8cm $$ and 
$$BB' = √3$$ 
So, by Pythagoras Theorem, the diagonal of this rectangle, 
$$B'C = \sqrt{(BB')^2+ (BC)^2}$$
$$ = \sqrt{(√3)^2+(8)^2}$$
$$ = \color{green}{\sqrt{67}} $$
Now, diagonal of rectangle $ \color{red}{ AA'B'B } $
$$AB' = \sqrt{(AA')^2+ (A'B')^2}$$
 $$AB'  = \sqrt{(√3)^2+10^2}$$
$$ =\color{red}{ \sqrt{103}} $$
So $$ AB' = \sqrt{103},$$
$$ AC = 6 ,$$
$$CB' = (√67) $$
Therefore, In ($3D$) $∆AB'C$,
We see that,
$$ (AB')^2=(\color{red}{√103})^2=103 $$
& $$ AC^2+(CB)^2 $$ is also
$$ = (\color{green}{√67})^2+(6)^2=67+36=103 $$
Since, square of larger side (hypotenuse) $ AB' $ is equal to sum of squares of other two sides, namely, $ AC $ & $ CB' $.
And therefore by  converse of Pythagoras Theorem , we have,
$$\angle ACB' = 90° $$
A: First, find the lengths of the triangle, but we do not need to find $AC$ because it is stated. To find $AB'$, find $AB$ first. $6^2+8^2=36+64=100=10^2$, so $AB$ is $10$cm. $10^2+\sqrt{3}^2=100+3=103=\sqrt{103}$, so $AB'$ is $\sqrt{103}$. $B'C$ is $\sqrt{8^2+3}=\sqrt{67}$ Taking these, we find that $AB'$ is longer than $AC$ and $AB'$, so $AC^2+AB'^2=AB'^2$ if $\angle ABC'$ is right (The Pythegorean theorem only works when the triangle is right). Taking these, we get: $\sqrt{67}^2+6^2=\sqrt{103}^2$, then $67+36=103$, then $103=103$. Therefore, $ABC'$ is a right-angle triangle.
Please note that $AA'$ has no unit, but I assumed it is $\sqrt3$.
