Let AD = h
Note that
AC = sqrt (h^2 + 2^2) = sqrt (h^2 + 4)
And
AB = sqrt( h^2 + 3^2) = sqrt ( h^2 + 9)
Using the Law of Cosines we have that
BC^2 = AC^2 + AB^2 - 2(AC * AB) cos (45°)
5^2 = (h^2 + 4) + (h^2 + 9) - 2 sqrt [ h^4 +13h^2 + 36] sqrt (2)/2
25 = 2 h^2 + 13 - sqrt [ h^4 + 13h^2 + 36] sqrt (2)
12 = 2h^2 - sqrt [ h^4 + 13h^2 + 36] sqrt (2)
12 - 2h^2 = - sqrt [ h^4 + 13h^2 +36] sqrt (2) square both sides
4h^4 - 48h^2 + 144 = (2)[h^4 + 13h^2 + 36]
4h^4 - 48h^2 + 144 = 2h^4 + 26h^2 + 72
2h^4 - 74h^2 + 72 = 0
h^4 - 37h + 36 = 0 factor
(h^2 - 1) ( h^2 - 36) = 0
(h + 1) ( h - 1) (h + 6 ) ( h - 6) = 0
The possible solutions are h = - 1, h = 1 , h = - 6 , h = 6
We can reject the negative answers
And h cannot be 1 because 180 - arctan (1/3) - arctan (1/2) = BAC = 180 - pi/4 = 135°
If h = 6 = AD then
180 - arctan (6/3) - arctan (6/2) = BAC = 180 - 3pi/4 = pi/ 4 = 45°
So....the area = (1/2) (BC) (AD) = (1/2)(5)(6) = 15 units^2