Question 1029869
Without loss of generality, situate triangle ABC on the Cartesian coordinate plane, with A(0,0), B(b,c), and C(a,0).  (The triangle has base on the positive x-axis.
With these points, it could be determined that X(a/2, 0) and Y({{{(a+2b)/3}}}, {{{(2c)/3}}}).

Now the line BX will have equation {{{y = (c/(b-a/2))(x-a/2)}}}, or {{{y = ((2c)/(2b-a))(x-a/2)}}}.

On the other hand, line AY will have equation {{{y = ((2c)/(a+2b))x}}}.

The intersection of lines BX and AY is point Z, which after solving is found out to be ({{{(a+2b)/4}}}, {{{c/2}}}).

From here it becomes easy:  {{{(AZ)/(ZY) = ((a+2b)/4)/((a+2b)/3 - (a+2b)/4) = ((a+2b)/4)/((a+2b)/12) = 3}}}.

The preceding calculation was based on ratios as determined from the x-coordinates.  A similar calculation based on ratios from the y-coordinates still yield the same value of 3.