Problem 11.4. A convex pentagon $A B C D E$ is such that $\angle A B C=128^{\circ}, \angle C D E=9^{\circ}$ $104^{\circ}, A B=B C, A E=E D$. How many degrees does the angle $A D B$ measure?

![](https://cdn.mathpix.com/cropped/2024_05_06_28b5ece3789ad9441c8ag-2.jpg?height=398&width=410&top_left_y=451&top_left_x=526)
Answer: 26.

![](https://cdn.mathpix.com/cropped/2024_05_06_28b5ece3789ad9441c8ag-2.jpg?height=470&width=517&top_left_y=968&top_left_x=468)

Fig. 14: to the solution of problem 11.4

Solution. The angles at the base of the isosceles triangle $ABC$ are equal to $\frac{1}{2}\left(180^{\circ}-128^{\circ}\right)=26^{\circ}$, and in the isosceles triangle $AED$ they are equal to $\frac{1}{2}\left(180^{\circ}-104^{\circ}\right)=38^{\circ}$.

Notice that $\angle ADC=90^{\circ}-38^{\circ}=52^{\circ}$, and the quadrilateral $ABCD$ is cyclic (Fig. 14), since $\angle ABC + \angle ADC = 128^{\circ} + 52^{\circ} = 180^{\circ}$. Therefore, $\angle ADB = \angle ACB = 26^{\circ}$.