You have shown no attempt of your own to do this so I have no idea what formulas or methods you might have to do such a problem and all I can do is say how I would do such a problem- with "analytic geometry". If we take the distance from B to D to be "L", we can set up a coordinate system so that B is at (0, 0) and D is at (L, 0). Then M is at (0, 20) and C is at (L, 60) so the line through M and C is given by y= (40/L)x+ 20. Also A is at (0, 40) and N is at (L, 30) so the line through A and N is given by y= (-10/L)x+ 40. Those lines intersect when (40/L)x+ 20= (-10/L)x+ 40 or (50/L)x= 20, x= (2/5)L. So h= (-10/L)(2/5)L+ 40= 36. As a check h= (40/L)(2/5)L+ 20= 36.

(Initially, I thought we could not do this without knowing the distance from B to D, L. But notice that L cancels out.)