Triangles The Invaders!

Since triangle $AJH$ is isosceles, $\angle JAH = \angle JHA = 12$ .

Since $\angle AJH$ and $\angle BJA$ are supplementary, $\angle BJA = 24$ ,

Since $\angle BJA$ and $\angle BAJ$ are complementary (they are the non-right angles of right triangle $BAJ$ ), $\angle BAJ = 66$ .

Since $\angle BAJ = \angle BAG$ is divided in $5$ equal parts, and $\angle DAJ$ is $\frac{3}{5}$ of $\angle BAJ$ , $\angle DAJ = 39.6$ .

$\angle DAH = \angle DAJ + \angle JAH = 39.6+12=\fbox{51.6}$ .

Above point J there are two more points, let's call point K and L where L is higher than K. We can see that angle ALH is about half of 156 which is 78. Because sum angle of triangle is always 180 and angle DAH and EHA is both same(in triangle ALH) so angle of DAH is 51