exam

angle ABC= angleACB   (Since AB=AC given)

let

ABC=ACB=x (say)

then

DAC=ABC+ACB=2x..(ext angle is equal to sum of int opp angles)

so

EAC=1/2[DAC]=x..(since AE  bisects DAC given)

So for lines BC  and AE, with AC as transversal,

EAC=ACB...(each equal to x)

so

BC parallel to AE..(alt int angles are equal)