relation between angle of incidence+angle of emergence=angle of prism+delta

pls prove it . and i request such informations to be added on meritnation .

i + e = A + δ

where

‘i' is the angle of incidence

‘e’ is the angle of emergence

‘A’ is the angle of prism

‘δ’ is the angle of deviation

$$\begin{gathered} \mathrm{Total} \mathrm{angle} \mathrm{of} \mathrm{deviation}=\mathrm{\delta }=\mathrm{angle} \mathrm{TQR}+\mathrm{angle} \mathrm{TRQ}={\mathrm{\delta }}_1+{\mathrm{\delta }}_2 \\ (\mathrm{Exterior} \mathrm{angle} \mathrm{of} \mathrm{triangle}=\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{interior} \mathrm{opposite} \mathrm{angles}) \\ \mathrm{\delta }=\mathrm{i}-{\mathrm{r}}_1+\mathrm{e}-{\mathrm{r}}_2 \\ =(\mathrm{i}+\mathrm{e})-({\mathrm{r}}_1+{\mathrm{r}}_2)-----\left(\mathrm{a}\right) \\ \mathrm{From} \mathrm{the} \mathrm{quadrilateral} \mathrm{AQOR}, \\ \mathrm{Angle} \mathrm{QAR}+\mathrm{Angle} \mathrm{ARO}+ \mathrm{Angle} \mathrm{ROQ}+\mathrm{Angle} \mathrm{OQA}={360}^0 \\ \mathrm{Here}, \mathrm{angle} \mathrm{ARO}=\mathrm{angle} \mathrm{OQA}={90}^0 \\ \mathrm{So}, \\ \mathrm{Angle} \mathrm{QAR}+\mathrm{Angle} \mathrm{ROQ}={180}^0--\left(\mathrm{i}\right) \\ \mathrm{From} \mathrm{the} \mathrm{triangle} \mathrm{ROQ} \\ {\mathrm{r}}_1+\mathrm{angle} \mathrm{ROQ}+{\mathrm{r}}_2={180}^0-----\left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \\ \mathrm{Angle} \mathrm{QAR}+\mathrm{Angle} \mathrm{ROQ}={\mathrm{r}}_1+\mathrm{angle} \mathrm{ROQ}+{\mathrm{r}}_2 \\ \mathrm{or} \mathrm{Angle} \mathrm{QAR}={\mathrm{r}}_1+{\mathrm{r}}_2 \\ \mathrm{Substituting} \mathrm{this} \mathrm{in} \left(\mathrm{a}\right), \mathrm{we} \mathrm{get} \\ \mathrm{\delta }=(\mathrm{i}+\mathrm{e})-({\mathrm{r}}_1+{\mathrm{r}}_2)=(\mathrm{i}+\mathrm{e})-\mathrm{Angle} \mathrm{QAR} \\ =(\mathrm{i}+\mathrm{e})-\mathrm{A} \\ \mathrm{or} (\mathrm{i}+\mathrm{e})=\mathrm{\delta }+\mathrm{A} \end{gathered}$$