Geometric problem based on angle bisectors I am not asking a question,i just want to conform,is my method of solving problem correct?

Given a triangle ABC.It is known that AB=4,AC=2,and BC=3.The bisector of angle A intersects the side BC at point K.The straight line passing through point B and being parallel to AC intersects the extension of angle bisector AK at point M.Find $(AM)^2$
My work:
$\angle CAK=\angle KMB$ (AC and BM are parallel)
$\angle ACK=\angle KBM$ (AC and BM are parallel) 
$\angle AKC=\angle BKM$  (vertically opposite angles)
so,$\bigtriangleup AKC and \bigtriangleup KBM$ are similar
In triangle ABM,angle BAK= angle KMB
therefore,AB=BM=4(side opposite to equal angles are equal.)
therefore,$\frac{AC}{BM}=\frac{KC}{KB}=\frac{AK}{KM}=\frac{1}{2}$
so,$KM=2 AK$
Also,$\frac{AB}{AC}=\frac{BK}{KC}=\frac{4}{2}=\frac{2}{1}$(angle bisector theorem)
BC=3(given)
therefore,$BK=\frac{2}{2+1}\times 3=2$,$KC=\frac{1}{2+1}\times 3=1$
For finding KM,i need to find AK,which i will find by using Stewart's theorem
$n=2,m=1,b=2,c=4,a=3$
$b^2n+c^2m=a(d^2+mn)$ Plugging values we get $d=AK=2$,
so,KM=2 AK=4,$KM^2=16$

You have a mistake 
$$b^2 n + c^2 m =a(d^2 + mn)$$
$$2^2 (2) + 4^2 (1) = 3(d^2 + 2)$$ 
$$8 + 16 = 3(d^2 + 2)$$ 
$$24 = 3(d^2 + 2) \Rightarrow d^2 = 6$$ 
Hence $KM^2 = (2AK)^2 = 4(AK^2) = 24$
