In ΔABC, AC > AB and AD is the bisector of ∠A show that y > x.

In ΔABC, since AC > AB 
∴ ∠B > ∠C ...... (i)
Also ∠1 = ∠2 ...... [given]
Now, in ΔABD, we have
∠1 + ∠B + ∠x = 180° ...... (ii)
And in ΔADC, we have
∠2 + ∠C + ∠y = 180° ...... (iii)
Comparing eqns (i) and (ii), we get 
∠1 + ∠B + ∠x = ∠2 + ∠C + ∠y
      ∠B + ∠x = ∠C + ∠y
Now since ∠B > ∠C ...... [proved above]
           .^.. ∠x < ∠y
           or ∠y > ∠x