triangle in parallelogram

[Kaza]

I don't know if my drawing is correctly. I think MN=1/3*BM and ∆DMP and ∆CMB are congruents (but I don't know how to demonstrate those) therefore DP=BC and BM=MP → BM=9 cm and BP=18 cm. Is that right?

 

[lev888]

You need to justify your statements. Why 1/3? Why congruent?
Read about similar triangles and various relationships between angles.

 

[Steven G]

Start labeling angles and sides and show us your updated diagram.

 

[pka]

View attachment 14203

I don't know if my drawing is correctly. I think MN=1/3*BM and ∆DMP and ∆CMB are congruents (but I don't know how to demonstrate those) therefore DP=BC and BM=MP → BM=9 cm and BP=18 cm. Is that right?

View attachment 14205

The triangles \(\displaystyle \Delta ANB \approx \Delta CNM\) by AAA.
Because \(\displaystyle AB=2\cdot CM\) corresponding parts, indeed \(\displaystyle MN=\tfrac{1}{3}BM\).
Can you show that \(\displaystyle \Delta DMP \approx \Delta CMB~?\)
Now finish!

 

[Kaza]

Because AD|| BC then <PDM=<MCB, <DMP=<BMC→ ∆DMP ~ ∆CMB → BM=MP→ BP=18 cm, q.e.d.
Thanks for help.