Perimeter of the triangle

Connect $\overline{AC}$ and draw a perpendicular line from $B$ down to $\overline{AC}$ and let the intersection be point $I$ . Because both $\angle EDB$ and $\angle DCA$ are $90 ^ \circ$ , $\overline{BI}$ would be parallel and equal to $\overline{DC}$ , $\overline{CI}$ would also be perpendicular and equal to $\overline{DB}$ , therefore $\overline{AI}\ = 5 - 4 = 1$ , and $\angle BIA = 90 ^ \circ$ .

According to the Pythagorean theorem:

$\overline{DC}$ = $\overline{BI}$ = $\sqrt{5^2 - 1^2} = \sqrt 24$ .

Then, according to the Tangent-secant theorem:

$\overline{DC}^2 = \overline{DH}\ \times \overline{DG}$ .

Let $\overline{DH} = x$

$x(10 + x) = 24$

$x^2 + 10x - 24 = 0$

$(x+12)(x-2)=0$

$x = -12$ or $2$ , $2$ being the only valid answer here.

Therefore, $P_{\triangle DBA} = 2 + 5 + 4 + 5 = 16$