Let $\vartriangle ABC$ be a triangle with $\angle BAC = 45^o, \angle BCA = 30^o$, and $AB = 1$. Point $D$ lies on segment $AC$ such that $AB = BD$. Find the square of the length of the common external tangent to the circumcircles of triangles $\vartriangle BDC$ and $\vartriangle ABC$.

Let's think step by step and output the final answer within \boxed{}.
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