help

Does this problem require trigonometry? This is very hard problem!            This problem should use similar triangles i believe

MN = x with triangle AMN similar to triangle ABX

Triangle ANC is proportional to AXC?

So set up proportion

\(\frac{x}{14}=\frac{5}{5+y}\)

With y being length of NX

How do we go from here????

If M is the mid point, then can't we just do \(\frac{x}{14}=\frac{1}{2}\)

x = 7           So MN = 7??????

Maybe not as difficult as it seems.....

Note that angles XNC  and ANC  are equal

And angles ACN and XCN  are equal

And NC = NC

Therefore, by ASA, triangles ANC and XNC are congruent

Then  AN  = XN  =  5     ....so.... AX  = 10

So ....N is the midpoint of  AX

Then.......if we connect MN.......this segment is paralell  to  BX because  MN splits the sides of triangle  ABX proportionally....that is   AM/BM  = AN/XN 

So  because MN is parallel to BX, then  triangle  AMN  is similar to triangle ABX  which implies that

MN / BX  =  AN / AX

MN  / 14   =  5 / 10

MN /14  = 1/2

MN  = (1/2) (14)    = 7   as found by CU!!!