How long of : BK = ?

Let the area of the parallelogram #ABCD# be #color(red)(24a)#,
Let #|ABC|# denote area of #DeltaABC#,
Let #G# be the midpoint of #CD#,
The three medians #AG,CF and DM# divide #DeltaACD# into 6 triangles with equal areas of #(24a)/(2xx6)=2a#, as shown in the figure,
#=> |MDA|=|MBC|=|MCD|=|MAB|=3xx2a=6a#
Given #AE=2EB#,
let #EB=s, => AE=2s#
Draw a line #MQ#, parallel to #CE#
as #M# is the midpoint of #AC#, and as #AE=2s#,
#=> AQ=(2s)/2=s, => QE=2s-s=s#
As #DeltaBKE and DeltaBMQ# is similar,
#=> (BK)/(BM)=(BE)/(BQ)=1/2#,
#=> BK=MK#
As #|CMB|=6a, and |CKB|:|CKM|=BK=MK=1:1#,
#=> |CKB|=|CKM|=3a#
#=> LK:KB=5a:3a#
#=> 10:KB=5:3#,
#=> KB=6#

Solution 2) :

The two diagonals bisect each other at #M#,
#=> DM=MB, AM=MC#,
As the two medians #DM and CF# intersect at #L#,
#=> L# is the centroid of #DeltaACD#,
#=> DM=3LM, => MB=3LM#
Given #AE=2EB#,
let #EB=s, => AE=2s#
Draw a line #MQ#, parallel to #CE#
as #M# is the midpoint of #AC#, and as #AE=2s#,
#=> AQ=(2s)/2=s, => QE=s, => BQ=2s#
As #DeltaBKE and DeltaBMQ# is similar,
#=> (BK)/(BM)=(BE)/(BQ)=1/2#,
#=> BK=MK=1/2MB=3/2LM#
#=> LM=2/3BK=2/3MK#
Given #LK=10#,
#LK=LM+MK=2/3MK+MK=5/3MK#
#=> 10=5/3MK, => MK=30/5=6#
#=> BK=MK=6#