The shape of the lot is a trapezoid or trapezium. If we call it ABCD labelling the points clockwise so that AB=54' on the road side, and DC=140' at the rear, with sides AD=171' and BC=124', we find the angle D is acute and angle C is obtuse, so the shape appears to be leaning over to the right. I assume AB is parallel to DC. If AD and BC are extended towards the roadside they will meet at a point we'll call X. We now have two similar triangles, XBA and XCD. We need first to find more information about the triangles before we can find out where the lot is 83' wide.
Between AB and DC we can draw a line YZ representing the 83' width and parallel to both AB and DC. That gives us another similar triangle XZY.
We have XA/XD=AB/DC, because the proportions of the corresponding sides of each similar triangle are the same. XA/(XA+171)=54/140. From this we get 140XA=54(XA+171)=54XA+9234. XA=9234/86=4617/43=107.372'. XA/XY=54/83 so 83XA=54XY, so XY=83*4617/(43*54)=165.035'. But XA+AY=XY, so AY=165.035-107.372=57.663'. AY is the distance from the front roadside measured along the long side (171') to the point where the width is 83'.
