AB=DC=a, AD=b:
sinx/AD = sin(54-x)/DC => sinx/b = sin(54-x)/a, or a/b=sin(54-x)/sinx
from triangle ABD, AB/AD=sin54/sin84 => a/b=sin54/sin84
so: sin(54-x)/sinx=sin54/sin84, or sin84sin54cosx - sin84cos54sinx = sin54sinx
=> tanx = sin84sin54/(sin54+sin84cos54), with calculator x = 30°
but, so far no clue how to get it from geometric method.