Object:
Factor a given rotation $ \rho_{Q,\theta} $ into the product of two reflections $ \rho = \sigma_a \sigma_m $, where $ m $ is a given mirror (through $ Q $ ).
Choose a gnomon. For example an F shaped concave hexagon. Make sure there is at least one point on the gnomon which is not hidden.
Give yourself an angle you can change, and later bisect.
Give yourself a center of rotation.
Rotate the F (pink to green) by the angle you made.
Given an arbitrary, point construct the first mirror $ m $. (Pink as well.)
Reflect the pink gnomon to the blue gnomon. Liberally wiggle the "givens" to see what depends on what.
Bisect the angle of rotation. You'll need half that angle.
Advance $ m $ to the after-line $ a $ by rotating $ m $ about $ Q $ by half the angle: $ a = \rho_{\theta/2}(m) $ .
Reflect the reflection in the blue mirror $ a $. BEHOLD !, it fits right on top of the rotated gnomon. You can follow the fate of point $ P $ to $ P' $ to $P''$.
Measure the angle $ PQP'' $ to see that it really is the angle we chose for our rotation. Wiggle the primitives vigorously to convince yourself of the truth of this theorem.
We called the mirror $ a $ for "after-mirror". Suppose we wanted to find a before-mirror $ b $ so that $ \rho =\sigma_m \sigma_b $. How would we do this? Submit a solution to this corollary consisting of the .ggb file and an (annotated) .png of your figure to the Moodle, when directed to do so.