Let $ABCD$ be the base of the pyramid, $V$ its apex, $VO$ its altitude and $M$, $N$ the midpoints of the edges $AD$, respectively $BC$. Denote $a$ the length of the edge $AB$. Faces $VAD$ and $VBC$ are perpendicular if and only if the triangle $VMN$ is right and isosceles, with sides $VM = VN = \frac{a\sqrt{2}}{2}$.

If $P$ is the foot of the perpendicular from $A$ on $VB$ (same as the foot of the perpendicular from $C$ on $VB$), then, computing in two ways the area of the triangle $VBC$ yields the equivalent condition $PC = PA = \frac{a\sqrt{6}}{3}$.

This is equivalent with $\sin \widehat{OPC} = \sqrt{3}/2$, that is $m(\widehat{OPC}) = 60^\circ$.

Since $\angle APC$ represents the angle of two consecutive lateral faces, this finishes the proof.