E is on the bisector of ∠ADC, EC⊥CD, EF⊥AD, so EC=EF, and because ED is the male side, △EDC and △EDF are congruent (HL).

Because e is the midpoint of CB, EC=EB, and it has been proved that EC=EF, so EB=EF, AE is the same side, and it is proved that △AEF and △ABF are congruent ∠EAB=∠EAF.

∠CED=35 degrees, then∠ ∠CDE=90 degrees -35 degrees =55 degrees.

Because DE is the bisector of ∠ADC, ∠ADC= 1 10 degrees.

∠C=∠B=90 degrees, and CD is parallel to AB, so∠ ADC +∠ DAB =180 degrees, so∠ ∠DAB=70 degrees.

It has been proved that AE is the bisector of ∠DAB, so ∠EAB=35 degrees.