From A to D, we should know the vertical distance from A to D, the horizontal distance from A to D, the width of the river, the vertical distance from A to the river and the vertical distance from D to the river.

As shown in the attached figure, let the horizontal distance from a to the intersection point b be x and the distance be y.

Y = √ (a 2+x 2)+c+√ [(d-x) 2+b 2], where a, b, c and d are known.

The question becomes: what is the value of x and the minimum value of y?

1, deformation:

y=√(a^2+x^2)+c+√[(x^2-2dx+d^2+b^2)，

2. Derive the above formula y'=dy/dx.

y'=2x/√(a^2+x^2)+2(x- 1)/√(x^2-2dx+d^2+b^2)，

3. Let y'=0 and get x0.

At this time, y(x0) is the extreme point, that is, the minimum value of y.

That is, when the horizontal distance between intersection B and starting point A is x0, and intersection B is between starting point A and destination point D, the distance is the shortest.