If I didn't understandd the question wrong. I find the solution. The four points will be the vertices of a trapezoid. Which has it's lengths a, a+m,a+m and a+3m, and its diagonal length will be a+2m .Lets say our trapeziod is ABCD so we can say AB is a, AD is a+m and AC is a+ 2m. BC is a+m and BD is a+2m. Last, CD is a+ 3m. Now let's find if it is possible or not. Let's say angle DAB is a . So DCB is 180 -a. From the cosinus theorem. (a+2m) ^2 = (a+m)^2 +a^2 -2.a.(a+m).cos a and again( a+2m)^2 = (a+3m)^2 + (a+m)^2 -(a+m).(a+3m)(-cos a) If you multiply the first equation with a+ 3m and the second equation with a+m. And sum up the two equations at the you will get the equation a^2 -am-3m^2=0 so for the a and m numbers satisfying the equation and the a>m>0 condition is apropriate from the side lengths of the trapezoid. For example (1, (square root(13) +1)/6 ) satisfies the condition for a and m. So the answer is YES.

Sorry the last equation should be a^2 +am -3m^2=0 but it doesn't change the answer.The example I give satisfies btw.