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 $ABC.$
A
30 30
B D C
Impossible
AD is a bisector
Angle CAD = 60/2 = 30
So...it can't also = 45
v
+843 
