In the quadrilateral $A B C D$, $A B=A C=D B$, and the diagonals are perpendicular to each other. Determine the sum of the angles $A C B$ and $A D B$.
Solution. Let the angle $ACB$ be $\alpha$, the angle $ADB$ be $\beta$, and the intersection point of the diagonals be $E$. In the right triangle $ECB$, $\angle EBC = 90^\circ - \alpha$, and in the right triangle $EDA$, $\angle EAD = 90^\circ - \beta$. The triangle $ACB$ is isosceles, so $\angle ABC = \alpha$, hence $\angle ABE = \alpha - (90^\circ - \alpha) = 2\alpha - 90^\circ$. Similarly, in the isosceles triangle $ABD$, $\angle BAE = \beta - (90^\circ - \beta) = 2\beta - 90^\circ$. In the right triangle $AEB$, we have $2\beta - 90^\circ + 2\alpha - 90^\circ = 90^\circ$, from which $\alpha + \beta = 135^\circ$.

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