3. In parallelogram $ABCD (AB \neq BC)$, two heights, $BH$ and $BK$, are drawn from the obtuse angle $B$ (the bases of the heights lie on the sides of the parallelogram and do not coincide with its vertices). Triangle $BHK$ turned out to be isosceles. Indicate all possible values of angle $BAD$.

(L. S. Korechkova)
Solution. Since the sides of the parallelogram are not equal, the heights are not equal either (for example, from the area formula), that is, $B H \neq B K$. This means that in triangle $B H K$, side $K H$ is equal to one of the other two. For definiteness, let's assume that $K H = B K$, and $H \in A D, K \in C D$.

Let the angle $K B H$ be $\alpha$ (it is acute because in triangle $B K H$ there are two such angles), then $\angle A B H = 90^{\circ} - \alpha$, and the required $\angle B A H = 90^{\circ} - \angle A B H = \alpha$.
![](https://cdn.mathpix.com/cropped/2024_05_06_7ac8905f0a14a1578722g-2.jpg?height=312&width=1446&top_left_y=1124&top_left_x=337)

Let's see for which $\alpha$ points $K$ and $H$ lie on sides $A D$ and $C D$. We will assume that we are given $\triangle B K H$ and we are constructing the parallelogram based on it. We need the segment $H K$ to be inside the right angles $B K D$ and $B H D$, that is, the angles $=180^{\circ} - 2 \alpha$ and $\angle B H K = \alpha$ must be acute. This is satisfied for $45^{\circ} < \alpha < 90^{\circ}$. (The diagram shows how the parallelogram will look for different values of $\alpha$.)

In addition, since $B H \neq B K$, then $\alpha \neq 60^{\circ}$.

Answer: any angle in the range $\left(45^{\circ} ; 90^{\circ}\right)$, except $60^{\circ}$.

Criteria. Only the correct answer -1 point.

The angle $60^{\circ}$ is not excluded - minus 1 point. Angles less than $45^{\circ}$ are not excluded - minus 2 points.