5. In $\triangle A B C$, it is known that $\angle A B C=44^{\circ}, D$ is a point on side $B C$, satisfying $D C=2 A B, \angle B A D=$ $24^{\circ}$. Then the size of $\angle A C B$ is $\qquad$.
5. $22^{\circ}$.

It is easy to know that $\angle A D C=44^{\circ}+24^{\circ}=68^{\circ}$.
As shown in Figure 3, draw ray

$A K$, such that $\angle D A K$
$=68^{\circ}$, and let it intersect $D C$
at point $K$.
Then $A K=D K$.

Since $\angle A K D=180^{\circ}-2 \times 68^{\circ}=\angle B$, we know that $A K=A B$.
Thus, $D K=A B$.
Since $D C=2 A B$, we have,
$$
K C=A B=A K, \angle C=\frac{1}{2} \angle A K D=22^{\circ}
$$