It is proved that the extension line from BP∑CD to ca after B is at P and H is a point on the extension line of bc.

Then < p = < PCD.

∠

pbc=∠dch

∠d=∠pba

∴△pab∽△cad

So ac/pa=ad/ba.

∫CD is the bisector of∠∠∠ BCA outer corner∠∠∠ ach.

∴∠pcd=∠dch=( 1/2)∠pch

So ∠ PBC = ∠ PCD = ∠ P

∴pc=bc=a

Pa = PC-AC = A-B。

∴ad/ba=ac/pa=b/(a-b)

ad=ba*[b/(a-b)]=bc/(a-b)