+118 
I have found that angle C is 75 degrees, but I don't know what to do next.
+279 Since ∠ABC=45∘ and AD is the angle bisector of ∠BAC, ∠BAD=∠CAD=260∘=30∘. Triangle ABD is a 30-60-90 triangle, so AB=2⋅AD=2⋅24=48.
Triangle ACD is a 30-60-90 triangle as well, so AC = sqrt(3)⋅AD = sqrt(3)*24.
Since AD is an angle bisector in triangle ABC,
[\frac{BD}{CD} = \frac{AB}{AC} = \frac{48}{24 \sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}.]
Then CD=3/(2*sqrt(3)) = 6.
The area of triangle ABC is then 1/2⋅BC⋅AD = 1/2⋅48⋅24 = 576.
+129845 
BAC = 60
So angle BAD =30 (since AD is a bisector)
Angle ABC =45
Therefore
DB / sin BAD = AD /sinBAD
DB / sin 30 = 24 / sin 45
DB / (1/2) = 24 /sin45
DB = (1/2) * 24 / [sqrt (2) / 2) = 12sqrt 2
Angle ADB = 180 - 45 - 30 = 105
So
AB / sin ADB = AD /sin ABC
AB / sin 105 = 24 / sqrt (2) /2
AB = 24sin105 /[sqrt (2) /2 ] = 12 + 12 sqrt (3)
AC / sin ABC = AB /sin ACB
AC / (sqrt (2) / 2) = (12 + 12 sqrt (3)) / sin 75
AC = (sqrt (2) / 2) (12 + 12sqrt (3)) / sin 75 = 24
[ABC ] = (1/2) AB * AC * sin 60
[ABC ] = (1/2) (12 + 12 sqrt (3)) *24 * sqrt (3) /2 = 72 (3 + sqrt 3) ≈ 340.7
