The sum of the ages of a man and his daughter is 44. Four years ago, the man was five times as old as his daughter. Find their present ages.

Let $M$ be the man's present age and $D$ be the daughter's present age.
According to the first statement, the sum of their ages is 44:

$$M + D = 44 \quad (1)$$

Four years ago, the man's age was $M - 4$.
Four years ago, the daughter's age was $D - 4$.
According to the second statement, the man was five times as old as his daughter four years ago:

$$M - 4 = 5(D - 4) \quad (2)$$

Expand the right side of equation (2):

$$M - 4 = 5D - 20$$

Add 4 to both sides:

$$M = 5D - 16 \quad (3)$$

Substitute $M = 5D - 16$ into $M + D = 44$:

$$(5D - 16) + D = 44$$ $$6D - 16 = 44$$

Add 16 to both sides:

$$6D = 44 + 16$$ $$6D = 60$$

Divide by 6:

$$D = \frac{60}{6}$$ $$D = 10$$

The daughter's present age is 10 years.

Using $M + D = 44$:

$$M + 10 = 44$$

Subtract 10 from both sides:

$$M = 44 - 10$$ $$M = 34$$

The man's present age is 34 years.

The man's present age is $\boxed{34 \text{ years}}$ and the daughter's present age is $\boxed{10 \text{ years}}$.