6. In a convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $E$. Given $\angle ABD=35^{\circ}, \angle ADB=20^{\circ}$, $\angle ACB=40^{\circ}, \angle ACD=70^{\circ}$. Then $\angle AEB=$ ).
(A) $60^{\circ}$
(B) $65^{\circ}$
(C) $70^{\circ}$
(D) $75^{\circ}$
6.D.

As shown in Figure 5, extend $AC$ to point $F$ such that $CF = CD$, and connect $FB$ and $FD$. It is easy to see that
$$
\begin{array}{l}
\angle AFD = \angle CFD \\
= \frac{1}{2} \angle ACD \\
= \frac{1}{2} \times 70^{\circ} = 35^{\circ} \\
= \angle ABD.
\end{array}
$$

Thus, points $A$, $B$, $F$, and $D$ are concyclic. At this point,
$$
\angle CFB = \angle AFB = \angle ADB = 20^{\circ} = \frac{1}{2} \angle ACB.
$$

Therefore, $CB = CF = CD$.
In isosceles $\triangle CBD$,
$$
\begin{array}{l}
\angle BCD = 40^{\circ} + 70^{\circ} = 110^{\circ}, \\
\angle BDC = \angle DBC = \frac{1}{2}(180^{\circ} - 110^{\circ}) = 35^{\circ}.
\end{array}
$$

Thus, $\angle AEB = \angle DBC + \angle ACB = 35^{\circ} + 40^{\circ} = 75^{\circ}$.