Let Haikal's three numbers be $a$, $b$, and $c$. According to the problem, we know the following sums:

$a + b = 23$ $b + c = 38$ $a + c = 45$

To find the sum of all three numbers, $a + b + c$, we can add the three given equations:

$(a + b) + (b + c) + (a + c) = 23 + 38 + 45$

Simplifying the left-hand side:

$a + b + b + c + a + c = 2a + 2b + 2c = 2(a + b + c)$

Now, calculate the right-hand side:

$23 + 38 + 45 = 106$

Thus, we have:

$2(a + b + c) = 106$

Dividing both sides by 2:

$a + b + c = \frac{106}{2} = 53$

So, the sum of all three of Haikal's numbers is $\boxed{53}$.

Would you like further details on how this was solved or have any other questions?

Here are 5 related questions:

1. What are the individual values of $a$, $b$, and $c$?
2. How would the solution change if one of the sums was different?
3. Can you express each number in terms of the others?
4. How would you solve this problem using matrices?
5. What if the problem involved four numbers and four sums?

Tip: When solving problems with sums of combinations, adding all the equations is a quick way to simplify and find the total.