We have a parallelogram $A B C D(|A B| \neq|B C|)$ with an internal angle of $72^{\circ}$ at vertex $A$. One of the vertices of this parallelogram is used to draw two lines, which divide the parallelogram into three isosceles triangles. Determine the sizes of the internal angles of these triangles.
For the internal angles of the given parallelogram $ABCD$, we have

$$
|\angle DAB|=|\angle BCD|=72^{\circ} \text { and } \quad|\angle ABC|=|\angle CDA|=180^{\circ}-72^{\circ}=108^{\circ} \text {. }
$$

Since $|AB| \neq|BC|$ and $|CD| \neq|DA|$, in both triangles $ABC$ and $CDA$, the obtuse angle is formed by two sides of different lengths, so they are not isosceles triangles. Therefore, the two mentioned lines must pass through either vertex $B$ or vertex $D$, and one of them is the line $BD$.

We can assume that the second line passes through vertex $B$ and intersects side $AD$ at point $P$ (in the pairs $B, D$ and $A, C$, the labeling of the vertices can be swapped). We will distinguish which of the sides $AB, AP, BP$ is the base of the isosceles triangle $ABP$. In each of the three cases, we can easily determine the internal angles of both isosceles triangles $ABP$ and $BPD$, and then calculate the angles of triangle $BCD$. Only in one case will we get an isosceles triangle:

![](https://cdn.mathpix.com/cropped/2024_04_17_d63c6598fbb9ece6fef7g-1.jpg?height=1010&width=1645&top_left_y=1411&top_left_x=228)

By appropriate division, we obtain isosceles triangles with the following internal angles: $72^{\circ}, 72^{\circ}, 36^{\circ} ; 36^{\circ}, 36^{\circ}, 108^{\circ} ; 72^{\circ}, 72^{\circ}, 36^{\circ}$.

ANOTHER SOLUTION. As in the first approach, we will only consider the situation where the parallelogram $ABCD$ is divided into isosceles triangles $ABP$, $BPD$, and $BCD$. We will distinguish which of the sides $BC, CD$ is the base of the third triangle (side $BD$ cannot be, since $|BC| \neq|CD|$):
![](https://cdn.mathpix.com/cropped/2024_04_17_d63c6598fbb9ece6fef7g-2.jpg?height=576&width=1240&top_left_y=518&top_left_x=430)

We will rule out the situation on the left image as follows: in the isosceles triangle $ABP$, the internal angle $ABP$ has one of the sizes $72^{\circ}, 54^{\circ}, 36^{\circ}$, so $|\angle ABC|>108^{\circ}$, which is a contradiction. On the right image, we will determine the remaining angles as follows: the isosceles triangle $ABP$ has an angle at vertex $P$ of $72^{\circ}, 54^{\circ}$, or $36^{\circ}$, while the isosceles triangle $BDP$ has an angle at vertex $P$ of $108^{\circ}, 72^{\circ}$, or $36^{\circ}$. The sum of these two angles with a common vertex $P$ is a right angle, so they are the angles $72^{\circ}$ and $108^{\circ}$.

[Justification for why the lines do not pass through points $A, C \ldots 2$ points, image with correctly filled angles ... 3 points, ruling out other possibilities ... 1 point.]