quick question, help!!!!!1

Here's one way to do this, tertre.....but maybe not the most elegant  !!!!

Let B  = (0,0)   C  = (8,0)

Since AM  is a median drawn to BC....then M  is  (4,0)

Now.....construct a circle with a radius of 5 centered at the origin

The equation of this circle is

x^2 + y^2  =  25      (1)

And construct a cirrcle with a radius of 4 centered at M

The equation of this circle is

(x - 4)^2 + y^2 =  16  (2)

Subtract  (2) from (1)   and we have that

x^2  - (x - 4)^2  =  9       simplify

x^2  - x^2 + 8x - 16   = 9

8x  = 25

x  = 25/8

This is the x coordinate of A

To find the y coordinate, we have

(25/8)^2 + y^2  =  25

625/64 + y^2  =  25

y^2   =  25  - 625/64

y^2  =  1600/64 - 625/64

y^2  = 975/64        take th positive root

 y = sqrt (975)/ 8

So  A   =  (25/8, sqrt (975/8)

So....using the distance formula  AC  =

sqrt  [ (8  - 25/8 )*2  + 975/64 ]  =

sqrt [ ( 39^2) / 64 + 975 / 64 ]  =

sqrt (39^2 + 975)  / 8  =

sqrt (2496)  / 8  =

sqrt  (2^6 * 3 * 13)  / 8  =

sqrt (64 * 3 * 13 ) / 8

8sqrt (39)  / 8

sqrt (39)

Here's a pic  :

Here is yet another simple method:

In the diagram of CPhill, triangle ABM is an isosceles triangle with legs of 4, 4, 5.
Using the Law of Cosines, angle ABM =51.3178, angle BAM =51.3178, angle AMB =77.3644
Triangle ABC now has legs of 5, 8, AC, with angle ABC =51.3178
Again, using Law of Cosines, AC =sqrt(39) = ~6.245....etc.

Thanks, tertre....I have a feeling that there is a better geometric way of doing this, but I don't know what it is....

Thanks to Guest for the good answer, too....the Law of Cosines works well!!!!