 Four points lying on one circle  Auxiliary circle $\quad]$

In a convex quadrilateral $A B C D$, it is known that $\angle A C B=25^{\circ}, \angle A C D=40^{\circ}$ and $\angle B A D=115^{\circ}$. Find the angle $A D B$
Quadrilateral $A B C D$ is convex, so ray $C A$ passes between the sides of angle $B C D$, hence,

$$
\angle B C D=\angle A C B+\angle A C D=25^{\circ}+40^{\circ}=65^{\circ} .
$$

Then

$\angle B C D+\angle B A D=65^{\circ}+115^{\circ}=180^{\circ}$,

therefore quadrilateral $A B C D$ is cyclic, i.e., points $A, B, C$ and $D$ lie on the same circle. Inscribed angles $A D B$ and $A C B$ in this circle subtend the same arc, therefore,

## Answer

$25^{\circ}$.

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