如图,△ABC中,AD是∠CAB的平分线,BD是△ABC的外角平分线,AD与BD交与于点D

∠D的度数为:70/2=35°.
设,∠CAD=∠DAB=∠1,
∠CBD=∠DBE=∠2.
∠ABC=180-(∠C+2∠1),
而,∠ABC=180-2∠2,
则有∠C+2∠1=2∠2,
∠2-∠1=∠C/2=35°.
∠D=180°-(∠1+∠ABC+∠2)
=180-∠1-∠2-(180-2∠2)
=∠2-∠1
=35°