How to solve finding quickest route physics problem

I am currently facing a physics problem that I am not sure of how to solve, and also, when I tried to put my system of equations into WolframAlpha, it didn't understand that it was a system of equations, and it returned an answer to one of the equations in the system.
The text of the problem is the following:

According to the diagram, a vehicle wants to go from point A to point C. The vehicle's speed on the road (AB) is constant v = 51 km/h, on the field (BC) is constant u = 28 km/h.
At which point B should the vehicle go off the road in order for it to get from point A to point C in the shortest time possible? (Or phrased differently, what is the distance of AB?) AD = d = 4 km, h = 3.9 km.

The diagram:

My attempt at the solution:

Data:
     v = 51 km/h  
     u = 28 km/h  
     d = 4 km  
     h = 3.9 km  

System of Equations:
     t(a,b,v,u) = a/v + b/u  
     b = Sqrt[h^2 + c^2]  
     d = a + c  
     x = u * cos[j]  
     y = u * sin[j]  
     x/c = y/h = u/b  
     h = 3.9  
     d = 4  
     v = 51  
     u = 28

As far as I can tell, our goal is to figure out what the minimum of function t is, but I'm not sure as to how to get it. I think I should use the similar triangles chb and xyu but again, I'm not sure about that.
I tried putting these equations into Wolfram Alpha, but it just returned the solution to b = Sqrt[h^2 + c^2] equation, and it also just returned that it's an equation of a circle, (since of course the equation of a circle is x^2+y^2=0, but this is of course Pythagorian Theorem.)

Could you please help me solve this little problem? Am I even on the right track to the solution? Also, how can I put this into WolframAlpha so that it understands properly?

Thank you for your time in advance. :)

This is a 1-DOF optimization problem. You have one variable you vary, the dimension c from 0 to d and you observe the total transit time t

t = a/v + b/u

Now it becomes a geometry problem, and not a physics problem as you need to express the distances a and b in terms of c and the other known quantities.

t = (d-c)/v + √(h^2+c^2)/u

If you plot this function of t in terms of c given v=51, u=28, d=4 and h=3.9 you will find that the minimum occurs somewhere near c≈3

You would have to use calculus to find the minimum exactly. Taking the derivative of time t vs distance c

t' = -1/v + c/(u*√(c^2+h^2))

and setting t' equal to zero, c*v - u*√(c^2+h^2)=0 produces the optimal distance c after re-arranging the terms and squaring both sides to get to

c^2/u^2 = c^2/v^2 + h^2/v^2

with solution

c = h * u/√(v^2-u^2) = 2.561 km