Problem with similarity of triangles and median.

Question

I have the next problem:

In the next image, MN // AB, PN = NC, QM = 8, BM = 6 and MC = 9. Calculate PM.

First I tried to find similarities in the triangles formed by the parallel sides, ABC and MNC.

So if NC = 9k, AN = 6K then AP = 3K. Then I tried to play with the angles, maybe angle ABC = 90 so there is a 6-8-10 triangle or angle PMC = 90 so MN = PN = NC but was unable to find anything else.

Thanks in advance.

Answer 1

I got what you mean. Note that there is no need to any extra assumption. The solution is as following:

Lets call

$$x=NC$$

$$\frac{AN}{NC}=\frac{6}{9}$$

so $$AN=\frac{2}{3}x$$

and $$PA=\frac{1}{3}x$$

There is another similarity between triangles PQA and PMN

Therefore:

$$\frac{PQ}{QM}=\frac{\frac{1}{3}x}{\frac{2}{3}x}$$

$$\frac{PQ}{8}=\frac{\frac{1}{3}x}{\frac{2}{3}x}$$

$$PQ=4$$