A geometry problem including Thales's theorem.

Question

Side $AB$ of parallelogram $ABCD$ is extended to a point $M$, so that $MD$ is sectioning side $BC$ in a point $N$, and that $BM = 2BN$. If $AD = 6$ cm. and $CD = 8$ cm., calculate in which ratio does point $N$ divide side $BC$. Now, I tried doing something using Thales' theorem, but I got that $12BN = 8BN$, so I am stuck with this. Any help?

Answer 1

Let BN = x . Therefore, BM = 2x

Now observe that $\triangle$MBN and $\triangle$MAD are similar.

Therefore : $\frac{MB}{MA}$ = $\frac{BN}{AD}$

$\frac{2x}{2x+8} = \frac{x}{6}$

So, x = 2

Required ratio BN : NC = x : 6 - x = 1 : 2

Answer 2

Hint: $\triangle MBN \sim MAD$. So $2 = \frac{BM}{BN} = \frac{MA}{AD}$.

Can you complete it from here to show that $BN = 2$?