$$\eqalign{
& {a \over b} = {c \over d} \Leftrightarrow ad = bc \Leftrightarrow {a \over c} = {b \over d} \cr
& {a \over c} = {b \over d} \Leftrightarrow {a \over c} + 1 = {b \over d} + 1 \Leftrightarrow {{a + c} \over c} = {{b + d} \over d} \Leftrightarrow c(b + d) = d(a + c) \Leftrightarrow {c \over d} = {{a + c} \over {b + d}} \cr
& {a \over c} = {b \over d} \Leftrightarrow {a \over c} - 1 = {b \over d} - 1 \Leftrightarrow {{a - c} \over c} = {{b - d} \over d} \Leftrightarrow c(b - d) = d(a - c) \Leftrightarrow {c \over d} = {{a - c} \over {b - d}} \cr
& \Rightarrow {a \over b} = {c \over d} = {{a + c} \over {b + d}} = {{a - c} \over {b - d}} \cr} $$