3. In a convex quadrilateral $A B C D: A B=A C=A D=B D$ and $\angle B A C=\angle C B D$. Find $\angle A C D$.
Answer: $70^{\circ}$.

Solution: Triangle $A B D$ is equilateral, so the angles $\angle A B D=\angle B D A=$ $\angle D A B=60^{\circ}$. Let $\angle B A C=\angle C B D=\alpha$, then $\angle A B C=60^{\circ}+\alpha . A B=A C$, thus $\angle A C B=\angle A B C=60^{\circ}+\alpha$. The sum of the angles in triangle $A B C$ is $\alpha+\left(60^{\circ}+\right.$ $\alpha)+\left(60^{\circ}+\alpha\right)=180^{\circ}, 3 \alpha=60^{\circ}, \alpha=20^{\circ}$. Angle $\quad \angle C A D=\angle B A D-\angle B A C=60^{\circ}-$ $20^{\circ}=40^{\circ}$. $A C=A D$, so $\angle A C D=\angle A D C=\beta$, the sum of the angles in triangle $A C D 40^{\circ}+2 \beta=180^{\circ}, 2 \beta=140^{\circ}, \angle A C D=\beta=70^{\circ}$.

Criteria: The angle $\angle B A C=\angle C B D=20^{\circ}$ is found - 4 points.