Using the notion of similar triangles, the relationship between the rigid bar deflection at points $\mathrm{A}$ and $\mathrm{D}$ can be expressed as:

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