From the question it is given that,
AD is bisector of ∠BAC
AB=6cm,AC=4cm and BD=3cm
Construction, from C draw a straight line CE parallel to DA and join AE
∠1=∠2 … [equation (i)]
By construction CE∥DE
So, ∠2=∠4 … [because alternate angles are equal] [equation (ii)]
Again by construction CE∥DE
∠1=∠3 … [because corresponding angles are equal] [equation (iii)]
By comparing equation (i), equation (ii) and equation(iii) we get,
∠3=∠4
So, AC=AE … [equation (iv)]
Now, consider the △BCE,
CE∥DE
BD/DC=AB/AE
BD/DC=AB/AC
3/DC=6/4
By cross multiplication we get,
3×4=6×DC
DC=(3×4)/6
DC=12/6
DC=2
Therefore, BC=BD+DC
=3+2
=5cm.