In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 60^\circ,$ $\angle CAD = 45^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$
let me elaborate on what BRAINBOLT said.
See, if D lies on the angle bisector of BAC, and angle BAC is 60 degrees, then angle CAD cannot be 45.
This is because since D lies on the bisector, CAD would therefore be half of BAC, which is 30.
So, this problem is invalid,
Thanks! :)