Working on the book: Lang, Serge & Murrow, Gene. "Geometry - Second Edition" (p. 46)
6. In the Figure 1.60, $d(P, Q) = d(R, S)$. Prove that $d(P, R) = d(Q, S)$.
The answer given in the Solutions manual, is:
Since $d(P,Q) = d(R,S)$ we can add $d(Q,R)$ to both sides of this equation and obtain $$ \begin{align} d(Q,R) &= d(P,Q) + d(Q,R) \tag{by SEG postulate}\\ &= d(R,S) + d(Q,R) \tag{by assumption}\\ &= d(Q,S) \tag{by SEG postulate} \end{align} $$
As I see it, $d(Q,R)$ is not equal to $d(P,Q) + d(Q,R)$. What's the equality the author is using to add $d(Q,R)$?