5. In $\triangle A B C$, the lengths of the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $c-a$ equals the height $h$ from vertex $A$ to side $AC$, then the value of $\sin \frac{C-A}{2}+\cos \frac{C+A}{2}$ is
A. 1
B. $\frac{1}{2}$
C. $\frac{1}{3}$
D. -1
$$
\begin{array}{l}
c-a=h=c \sin A \Rightarrow \sin C-\sin A=\sin C \sin A \\
\Rightarrow 2 \cos \frac{C+A}{2} \sin \frac{C-A}{2}=-\frac{1}{2}[\cos (C+A)-\cos (C-A)] \\
=\cos ^{2} \frac{C-A}{2}-\cos ^{2} \frac{C+A}{2}=1-\sin ^{2} \frac{C-A}{2}-\cos ^{2} \frac{C+A}{2} \\
\Rightarrow \sin ^{2} \frac{C-A}{2}+\cos ^{2} \frac{C+A}{2}+2 \cos \frac{C+A}{2} \sin \frac{C-A}{2}=\left(\sin \frac{C-A}{2}+\cos \frac{C+A}{2}\right)^{2} \\
=1 \Rightarrow \sin \frac{C-A}{2}+\cos \frac{C+A}{2}=1, \text { so choose } A .
\end{array}
$$