Measure of $\angle D =?$

$\angle EGC=70^\circ\implies\angle CAB=70^\circ$ because $AB$ and $EH$ are parallel.

$AB=BC\implies \angle ACB=70^\circ$ as well, and therefore also $\angle DCG=70^\circ$

Let $\angle CDF=x$ . Then because $BD$ is parallel to $FH$ the $\angle EFH=180^\circ-x$ .

$\angle FEH=\frac12(180-\angle EFH)=\frac x2$

$\angle DEG=180^\circ-\angle EFH=180^\circ-\frac x2$

The sum of the angles in rectangle $DEGC$ is

$360^\circ=x+70^\circ+70^\circ+180^\circ-\frac x2$

From that $x=\boxed{80^\circ}$

$\angle EGC=70°$ , so ( $AB$ and $EH$ are parallel) $\angle CAB=70°$ .

$AB=BC$ , so $\angle ACB=70°=\angle DCG$

If $M$ is the midpoint of $BD$ and $EH$ (because they can't be parallel), then $\angle CMG=180°-2*70°=40°$ . Since $BD$ and $FH$ are parallel, $\angle EHF=40°$ . So $\angle CDE=360°-\angle DCG-\angle CGE-\angle GED=360°-70°-70°-140°=\boxed{80°}$