Question

How long of : BK = ?

Answer 1

`BK=x=6`

Explanation:

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`

Answer 2

`BK=6`

Explanation:

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`