∠1=∠2 (Given)

Draw a line EC∥AD

AC bisects them

∴∠2=∠3 (by alternate angle) ………………. (i)

∠1=∠4 (corresponding angle) ……………..(ii)

∠1=∠2 (given)

From equ (i) and equ (ii)

∠3=∠4

or AE=AC ……………..(III)

Now ,⊿ BCE

BD/DC=BA/AE ( BY PROPORTIONALITY THEORAM)

BD/DC=AB/AC ( ∵ BA=AB AND AE=AC from equ (iii))

Hence AB/AC=BA/DC Proved