Note that minor arcs
AD + AB + BC + CD = 360° or....put another way.....
AD + 3BC = 360 → AD = 360 - 3BC (1)
Also
Angle BEC = (1/2) ( BC - AD) sub (1) into this
40 = (1/2) [ BC - (360 - 3BC) ] multiply through by 2 and simplify
80 = 4 BC - 360 add 360 to both sides
440 = 4 BC divide both sides by 4
110° = BC
So, since AB = BC = CD.....then major arc ACD = 330°
And arc AD must = 30°
And since angle ACD is an inscribed angle intercepting this arc...
Its measure is (1/2) arc AD = 15°
