The Three Perpendiculars Theorem

Fix a point $A$ on $d$ outside $(\alpha)$, and let $H$ be the foot of the perpendicular from $A$ to $(\alpha)$, so $AH \perp (\alpha)$; by definition of orthogonal projection, $H$ lies on $d_1$. Because $AH$ is perpendicular to the entire plane $(\alpha)$, it is perpendicular to the line $a$ inside that plane: $AH \perp a$. Meanwhile $d$, $d_1$ and $AH$ all lie in one vertical plane.

($\Rightarrow$) Suppose $a \perp d$. Then $a$ is perpendicular to both $d$ and $AH$, which are two intersecting lines of that vertical plane. By the line–plane perpendicularity criterion, $a$ is perpendicular to the whole vertical plane, and therefore to $d_1$, which lies in it; hence $a \perp d_1$.

($\Leftarrow$) Conversely, suppose $a \perp d_1$. Then $a$ is perpendicular to both $d_1$ and $AH$, again two intersecting lines of the same vertical plane. The same criterion makes $a$ perpendicular to that vertical plane, and therefore to $d$; hence $a \perp d$.