I have been thinking about this problem, but I can't get anywhere.
The figure shows a circle with A as its center and has a radius of 5.
EC and EF are both tangent to the circle, and DG is a secant line that passes through A and intersects with the circle at G.
∠EDB=90∘, and DB=4.
Find the perimeter of triangle DBA.
CA will be perpendicular to CDE (radius to tangent).
DB || CA (both perpendicular to EC).
Find point X on CA such that DX || BA.
XA = 4 (DBAX is a parallelogram).
CX = 1.
DX = 5.
Triangle(DCX) is a right triangle ---> CX2 + CD2 = DX2 ---> 12 + CD2 = 52
---> CD = sqrt(24)
Triangle(DCA) is a right triangle ---> DC2 + CA2 = DA2 ---> [sqrt(24)]2 + 52 = DA2
---> DA = 7
Now, you can find the perimeter of triangle(DBA).