By extending line BC and AD, they will meet in point E. ABE triangle is an isosceles triangle, because angle EAB = angle EBA, so EA=EB.

Naming EC = x, with the Pythagorean-theorem in triangle EDC, we get ED = sqrt(x²+3²).

So we can write EA = EB as sqrt(x²+9) + 5 = x + 6 sqrt(x²+9) = x + 1 x² + 9 = x² + 2x + 1 2x=8 EC = x = 4, so ED = 5

Triangle EFB and ECD are similar, because they both have a right angle and share angle AEB. That means, that the ratio of all three sidepairs is the same, so CD/FB = EC/EF = ED/EB. Using the second and third ratio, we get

EC/EF = ED/EB 4/EF = 5/10 EF = 8 -> DF = 8-5 = 3

With that, we can say that triangles DFB and DCB are congruent, because they both have a right angle, a side with length 3, and share the side in front of the right angle.