Problem 5. In quadrilateral $A B C D$, it is known that $A B=$ $=B C=C D, \angle A=70^{\circ}$ and $\angle B=100^{\circ}$. What can the angles $C$ and $D$ be equal to? $\quad[8$ points] (M.A.Volchkevich)
Answer: $60^{\circ}$ and $130^{\circ}$ or $140^{\circ}$ and $50^{\circ}$.

First Solution. Draw segment $B E$ such that point $E$ lies on $A D$, and angle $A B E$ is $40^{\circ}$. Then $\angle A E B = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ}$, hence triangle $A B E$ is isosceles, $A B = B E$. Consider triangle $B C E$: $\angle C B E = 100^{\circ} - 40^{\circ} = 60^{\circ}$ and $B E = A B = B C$, thus triangle $B C E$ is equilateral, and $C E = B C = A B$. This means that quadrilateral $A B C E$ satisfies the condition, and one possible answer is that angle $C$ of such a quadrilateral is $60^{\circ}$, and the remaining angle is $\angle A E B + \angle B E C = 70^{\circ} + 60^{\circ} = 130^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_208a41587bfd2c790cf7g-3.jpg?height=331&width=547&top_left_y=1490&top_left_x=451)

Notice that for any point $D^{\prime}$ on segment $A E$, $C D^{\prime} > C E = B C$ (since $C D^{\prime}$ is the largest side in the obtuse triangle $C E D^{\prime}$). Let point $D$ lie on ray $A E$ beyond point $E$, and $C D = B C = C E$. Then $\angle C E D = 180^{\circ} - \angle A E B - \angle B E C = 180^{\circ} - 70^{\circ} - 60^{\circ} = 50^{\circ}$, and since $C E = C D$, $\angle C D E = \angle C E D = 50^{\circ}$, thus $\angle E C D = 180^{\circ} - 2 \cdot 50^{\circ} = 80^{\circ}$ and $\angle B C D = 60^{\circ} + 80^{\circ} = 140^{\circ}$.

Second Solution. In the isosceles triangle $A B C$, angle $A B C$ is $100^{\circ}$, so $\angle B A C = \angle A C B = 40^{\circ}$, and then $\angle C A D = 70^{\circ} - 40^{\circ} = 30^{\circ}$. Mark the midpoint $M$ of segment $A C$ and the foot $P$ of the perpendicular dropped from point $C$ to line $A D$. Then, in the right triangle $A C P$, the side opposite the $30^{\circ}$ angle is $C P$, so $C P = \frac{1}{2} A C = C M$. Therefore, right triangles $B C M$ and $D C P$ are congruent by the hypotenuse and one leg, so $\angle C D P = \angle M B C = 50^{\circ}$, i.e., depending on whether point $P$ lies inside segment $A D$ or outside, either angle $A D C$ or its adjacent angle is $50^{\circ}$. Accordingly, in the first case, $\angle A D C = 50^{\circ}$ and $\angle B C D = 140^{\circ}$, and in the second case, $\angle A D^{\prime} C = 130^{\circ}$ and $\angle B C D^{\prime} = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_208a41587bfd2c790cf7g-4.jpg?height=327&width=549&top_left_y=1027&top_left_x=452)