Answer please. Need help

Step One
Show that ΔADC ≡  ΔACB are similar Maybe ≡ this might mean congruent, I'm not sure. I want similarity.

A is common to both triangles.
<ADC = <ACB Both angles are right angles.
Conclusion
ΔADC is similar to ΔACB    Angle Angle theorem which is enough to declare similarity.

Step Two
Find AB
Set up a proportion.
AD/AC = AC/AB corresponding parts of similar triangles bear the same ratio.

Substitute and solve for AB
2/5 = 5/AB Cross multiply
2AB = 25 Divide by 2
AB = 13/2
AB = 6.5

Step Three
Now you need the height of the triangle (CD)
That's just a^2 + b^2 = c^2

a = 2
c = 5
b = ???
2^2 + b^2 = 5^2
4 + b^2 = 25 Subtract 4 from both sides.
b^2 = 25 - 4
b^2 = 21
b = sqrt(21)

Step 4
Find the area.
A = 1/2 b*h
h = sqrt(21)
b = 6.5

Area = 1/2 6.5 * sqrt(21)
Area = 14.89