$AD = ({\hat A_1} = {\hat A_2})\,,\,AB = AD$ = فرض

اندازهٔ ${D_1}$ = حکم

$AD = AB \to {\hat D_1} = B$ چون

${\hat A_1} + \hat B + {\hat D_1} \to 180 \to 2\hat B + {\hat A_1} = {180^ \circ }$ چون مجموع زوایای داخلی برابر

$ \to \underbrace {\hat A}_{{A_1} + {A_2}} + \hat B + \hat C = {180^ \circ } \to 2{\hat A_1} + \hat B + {30^ \circ } = 180 \to 2{\hat A_1} + \hat B + {150^ \circ }$ و همچنین در مثلث  $\mathop {ABC}\limits^\Delta  $

اگر ${\hat A_1} = x$ و $\hat B = y$ باشد.

$\begin{gathered}
  \left\{ \begin{gathered}
  x + 2y = 180 \hfill \\
  2x + y = 150 \hfill \\ 
\end{gathered}  \right. \Rightarrow \underline {\left\{ \begin{gathered}
  x + 2y = 180 \hfill \\
   - 4x - 2y =  - 300 \hfill \\ 
\end{gathered}  \right.}  \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3x + 0 =  - 120\, \to \,x = {40^ \circ } = {A_1} \hfill \\ 
\end{gathered} $

$40 + 2y = 180 \to 2y = 140 \to y = 70 = B \to {D_1} = 70$