Reflection of a Plane Wave by Huygens' Construction

Consider a plane wavefront $AB$ travelling through a medium and about to strike a plane reflecting surface $MN$, making an angle of incidence $i$ with the normal to the surface at the point where it first touches the surface. Let $\tau$ be the time taken for the point $B$ of the wavefront to travel from $B$ to a point $C$ on the reflecting surface, so that $BC=v\tau$, where $v$ is the speed of the wave in the medium. By Huygens' principle, during this very same interval $\tau$, the point $A$ of the wavefront -- which reached the surface earlier -- has itself been acting as a source of a secondary wavelet spreading back into the same medium (since the wave is reflected, not transmitted), and this secondary wavelet has by now grown to a hemisphere of radius $AE=v\tau$ centred on $A$. The new, reflected wavefront is the common tangent $CE$ drawn from $C$ to this secondary wavelet.

Since $AC$ is the hypotenuse common to both right triangles $ABC$ (right-angled at $B$) and $AEC$ (right-angled at $E$), and $BC=AE=v\tau$, the two triangles are congruent by the RHS (right angle-hypotenuse-side) test. Congruence gives $\angle BAC=\angle ACE$ directly -- and since $\angle BAC$ is the angle the incident wavefront makes with the surface (equal to the angle of incidence $i$, measured from the normal) and $\angle ACE$ is the corresponding angle for the reflected wavefront (equal to the angle of reflection $r$), this proves

$i=r$ …