1956 AMC 12 Problem 50

Scale $CA=CB$ to $1,$ and place $C=(0,0),$ $B=(1,0),$ $A=(\cos\theta,\sin\theta)$

Because the square is outside the triangle, take $D=(0,-1)$ and compare vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$

Scale the equal sides to $1$ and place $$\begin{aligned} C&=(0,0),\qquad B=(1,0),\\ A&=(\cos\theta,\sin\theta). \end{aligned}$$ Since square $BCDE$ is constructed away from the triangle, $D=(0,-1).$ Put $c=\cos\theta$ and $s=\sin\theta.$ Then $$\begin{aligned} \overrightarrow{AB}&=(1-c,-s),\\ \overrightarrow{AD}&=(-c,-1-s). \end{aligned}$$ Their dot product is $1+s-c,$ while the absolute value of their two-dimensional cross product is also $1+s-c.$ Therefore $$\tan\angle DAB=1,$$ so $\angle DAB=45^\circ,$ independent of $\theta$ and hence independent of the triangle’s shape.

Thus, the correct answer is B.