In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 45^\circ,$ $\angle ABC = 45^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$
45 A
24
C D B 45
Angle CAD =22.5 Angle CDA = 67.5 Angle ACD = 90
AC / sin (CDA) = AD / sin (ACD)
AC/ sin (67.5) = 24 / 1
AC = 24sin (67.5)
BAC = 45 = ABC
So BC = AC
Area = (1/2)(BC * AC) = (1/2)[ 24 sin (67.5)] ^2 ≈ 245.8
+1234 
