jmc

First, we build a diagram:



We have $\angle AHB=\angle DHE$, and from quadrilateral $CDHE$, we have $$\begin{array}{rl}\angle DHE& =360^{\circ }-\angle HEC-\angle ECD-\angle CDH\\ & =360^{\circ }-90^{\circ }-\angle ACB-90^{\circ }\\ & =180^{\circ }-\angle ACB.\end{array}$$From triangle $ABC$, we have $180^{\circ }-\angle ACB=\angle BAC+\angle ABC=54^{\circ }+52^{\circ }=\boxed{106^{\circ }}$.