geometry

\(CA=\sqrt{20^2+16^2}=\sqrt{656}\)

\(BD^2+20^2=BA^2\)

\(AC^2+BA^2=(DC+BD)^2\) =>\(656+(BD^2+20^2)=(16+BD)^2\)

\(656+BD^2+400=256+32BD+BD^2\)

\(1056-256=800=32BD\)

\(BD=25\)

Thanks, James

Another  method

We  have  the relationship that

BD / AD  = AD  /  DC

BD / 20  =  20 / 16           cross-multiply

16 BD  =  20 * 20

16 BD  =  400

BD  =  400  /16     =   25

BD = 20(20 / 16)