One approach is to use similar triangles. A sketch will do: there's no need to draw to scale. Draw a tall thin isosceles triangle ABC, where AB=AC, and the base BC represents the rear of the lot 58 feet. Draw a line WX parallel to BC to represent the 51 foot front of the lot. The quadrilateral WXCB is a trapezoid or trapezium representing the lot. WB=XC represents 125 feet. We have two similar triangles ABC and AWX. Now draw a line YZ parallel to WX and BC close to WX, between WX and BC. YZ represents the width 54 feet. We now have a third similar triangle AZY.
AB/BC=AW/WX=AY/YZ because of similar triangles. We have some values we can put into this:
AB/58=AW/51=AY/54. AB=AW+WB=AW+125. Let AW=x, then we can write (x+125)/58=x/51; cross-multiplying we get: 51x+6375=58x, so 7x=6375, x=6375/7. Let WY=y, the distance along the side of the lot where the width is 54 feet. x/51=AY/54=(AW+WY)/54=(x+y)/54. Cross-multiplying we get: 54x=51(x+y); 54x=51x+51y; 3x=51y, so y=3x/51=x/17. We already know x so y=6375/(17*7)=375/7=53.57 feet. So if we measure 53.57 feet from the front of the lot along the 125-ft side the width at that point will be 54 feet.