solution: Since $\angle BCA = 40^\circ$ and $\triangle ADC$ is isosceles with $AD=DC,$ we know $\angle DAC=\angle ACD=40^\circ.$
  
  Since the sum of the angles in a triangle is $180^\circ,$ we have \begin{align*}
  \angle ADC &= 180^\circ - \angle DAC - \angle ACD \\
  &= 180^\circ - 40^\circ - 40^\circ \\
  &= 100^\circ.
  \end{align*}Since $\angle ADB$ and $\angle ADC$ are supplementary, we have  \begin{align*}
  \angle ADB &= 180^\circ - \angle ADC \\
  &= 180^\circ - 100^\circ \\
  &= 80^\circ.
  \end{align*}Since $\triangle ADB$ is isosceles with $AD=DB,$ we have $\angle BAD = \angle ABD.$ Thus, \begin{align*}
  \angle BAD &= \frac{1}{2}(180^\circ - \angle ADB) \\
  &= \frac{1}{2}(180^\circ - 80^\circ) \\
  &= \frac{1}{2}(100^\circ) \\
  &= 50^\circ.
  \end{align*}Therefore,  \begin{align*}
  \angle BAC &= \angle BAD + \angle DAC \\
  &= 50^\circ+40^\circ \\
  &= \boxed{90^\circ}.
  \end{align*}
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  In the diagram, $AD=BD=CD$ and $\angle BCA = 40^\circ.$ What is the measure of $\angle BAC?$
  
  [asy]
  draw((0,0)--(10,0)--(8.2635,9.8481)--cycle,black+linewidth(1));
  draw((10,0)--(20,0)--(8.2635,9.8481),black+linewidth(1));
  draw((5,-0.5)--(5,0.5),black+linewidth(1));
  draw((15,-0.5)--(15,0.5),black+linewidth(1));
  draw((8.6318,4.8359)--(9.6317,5.0122),black+linewidth(1));
  label("$A$",(8.2635,9.8481),N);
  label("$B$",(0,0),SW);
  label("$C$",(20,0),SE);
  label("$D$",(10,0),S);
  [/asy]
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 90^\circ
extra_info.question: In the diagram, $AD=BD=CD$ and $\angle BCA = 40^\circ.$ What is the measure of $\angle BAC?$
  
  [asy]
  draw((0,0)--(10,0)--(8.2635,9.8481)--cycle,black+linewidth(1));
  draw((10,0)--(20,0)--(8.2635,9.8481),black+linewidth(1));
  draw((5,-0.5)--(5,0.5),black+linewidth(1));
  draw((15,-0.5)--(15,0.5),black+linewidth(1));
  draw((8.6318,4.8359)--(9.6317,5.0122),black+linewidth(1));
  label("$A$",(8.2635,9.8481),N);
  label("$B$",(0,0),SW);
  label("$C$",(20,0),SE);
  label("$D$",(10,0),S);
  [/asy]
extra_info.split: test