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.

Guest Jun 22, 2020

#1**+1 **

Draw CA.

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 ---> CX^{2} + CD^{2} = DX^{2} ---> 1^{2} + CD^{2} = 5^{2}

---> CD = sqrt(24)

Triangle(DCA) is a right triangle ---> DC^{2} + CA^{2} = DA^{2} ---> [sqrt(24)]^{2} + 5^{2} = DA^{2}

---> DA = 7

Now, you can find the perimeter of triangle(DBA).

geno3141 Jun 22, 2020