= Solution
The relevant planes contain $[100]$, so their normals lie in the $bc$ plane. The symmetry operations can be checked directly on the height-labelled molecules:
* \b[<Glide planes> of type $c$, parallel to $(010)$ at $y/b=1/4,3/4$], for example
$$
(x,y,z)\longmapsto(x,1/2-y,z+1/2).
$$
Reflection reverses the bond slope, and the $c/2$ translation carries A to B with the correct height.
* \b[<Glide planes> of type $a$, parallel to $(001)$ at $z/c=1/4,3/4$], for example
$$
(x,y,z)\longmapsto(x+1/2,y,1/2-z).
$$
This again exchanges A and B. Combining it with the C-centering translation gives the equivalent description with a $b/2$ glide in the same plane.
\b[There are no ordinary mirror planes containing $[100]$ for the illustrated generic bond inclination.] Reflection alone would reverse a bond slope without the necessary accompanying translation. A mirror perpendicular to $[100]$ is a different symmetry and is outside the requested zone.
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