Substitute

$$\begin{gathered} {\mathrm{v}}_{\mathrm{A}}={\mathrm{\delta }}_1 \\ {\mathrm{v}}_{\mathrm{D}}=-{\mathrm{\delta }}_2{ \\ }_{ \\ } \end{gathered}$$

into the equation from the rigid bar geometry

$$\begin{gathered} \frac{{\mathrm{v}}_{\mathrm{A}}}{750 \mathrm{mm}}=\frac{{\mathrm{v}}_{\mathrm{D}}}{300 \mathrm{mm}}{ \\ }_{ \\ } \end{gathered}$$

to obtain the geometry of deformation equation:

$$\begin{gathered} \frac{{\mathrm{\delta }}_1}{750 \mathrm{mm}}=\frac{-{\mathrm{\delta }}_2}{300 \mathrm{mm}}{ \\ }_{ \\ } \end{gathered}$$