In a parallelogram $A B C D$, the perimeter is $35 \mathrm{~cm}$, and the bisectors of the interior angles at vertices $A$ and $B$ intersect at a point $O$ outside the quadrilateral. Calculate the sides of the parallelogram, given that the side $D C$ bisects the segment $A O$ into two equal parts.
Point $O$ is always between the lines $A D$ and $B C$, and according to the condition, it is outside the parallelogram. This can only be the case if the intersection points of the $D C$ side with the segment $A O$ at $E$ and with the segment $B O$ at $F$ lie on the segment $D C$, with the order of points being $D, E, F, C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_947a47522227a3bba5edg-1.jpg?height=372&width=659&top_left_y=302&top_left_x=733)

According to the condition, $A E = E O$, and $E F$, as a part of the side $D C$, is parallel to $A B$, thus it is the midline of the triangle $A B O$: $E F = A B / 2 = C D / 2$.

The triangles $A D E$ and $B C F$ are isosceles because the angle $D A E$ is half of the angle $D A B$, and the angle $D E A$ is the vertical angle of the angle $O E C$, which is congruent to the angle $O A B$, the other half of the angle $D A B$. Similarly,

$$
\angle C F B = \angle O F D = \angle O B A = \frac{1}{2} \angle C B A = \angle C B F.
$$

The legs of the triangles are $D E = D A = C B = C F$. We also know that $D E + C F = C D - E F = \frac{1}{2} C D$, so $C D = 4 A D$, and the perimeter of the parallelogram is 10 times $A D$. Therefore, the length of $A D$ is $3.5 \, \text{cm}$, and the length of $C D$ is $14 \, \text{cm}$.

Borbála Horváthy (Esztergom, Bottyán J. Mechanical Engineering High School I. o. t.)

Note. The conditions determine the lengths of the sides, but the angle of the parallelogram can be arbitrary.