solution: By Pythagoras, $\angle C = 90^\circ$.  Triangles $ACD$ and $ABC$ are similar, so \[CD = BC \cdot \frac{AC}{AB} = 15 \cdot \frac{8}{17} = \frac{120}{17},\]and \[AD = AC \cdot \frac{AC}{AB} = 8 \cdot \frac{8}{17} = \frac{64}{17}.\][asy]
  unitsize(0.4 cm);
  
  pair A, B, C, D;
  
  A = (0,8);
  B = (15,0);
  C = (0,0);
  D = (C + reflect(A,B)*(C))/2;
  
  draw(A--B--C--cycle);
  draw(C--D);
  
  label("$A$", A, NW);
  label("$B$", B, SE);
  label("$C$", C, SW);
  label("$D$", D, NE);
  [/asy]
  
  Therefore, the area of triangle $ACD$ is \[\frac{1}{2} \cdot AD \cdot CD = \frac{1}{2} \cdot \frac{64}{17} \cdot \frac{120}{17} = \boxed{\frac{3840}{289}}.\]
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  In triangle $ABC$, $AB = 17$, $AC = 8$, and $BC = 15$.  Let $D$ be the foot of the altitude from $C$ to $AB$.  Find the area of triangle $ACD$.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: \frac{3840}{289}
extra_info.question: In triangle $ABC$, $AB = 17$, $AC = 8$, and $BC = 15$.  Let $D$ be the foot of the altitude from $C$ to $AB$.  Find the area of triangle $ACD$.
extra_info.split: test