First I am going to take exception to the last sentence, "Perpendicular rays DO NOT refract (reflection only). Perpendicular rays reflect and transmit, though the transmitted rays don't change their angle (or direction). See Snell's law.
Second: It is true that the transmitted waves change their wavelength $\lambda$ and speed $v$, $\lambda_2=\nu_1*\lambda_1/\nu_2$, and $v_2=\nu_1*v_1/\nu_2$, but since frequency=speed/wavelength, $\nu_2=\nu_1$. (n= refraction index).
But looking on this from the point of view of $E=h\nu$, and getting the outcome of $2h\nu$ is an interesting question. That view is a photon view, and a single photon will just take one of the two paths, reflected or refracted, much like in two slit interference, or an image being formed. For the photon view to work you need to look at photon statistics. I once did a thin film coating transmission and reflection calculation doing photon statistics at each interface along with interference for each beam.
So taking many photons and the probabilities of being reflected or refracted (or?) I am going to deviate a little from some texts and say the incident power = refracted power + reflected power + absorbed power + scattered power does hold.
Another thing to consider is the intensity (power/area) of the refracted beam. Due to refraction it does not have the same cross section of the incident beam. All of this is calculated within the Fresnel equations for the electric and magnetic fields for reflection and refraction in any E&M textbook.