Problem 7.6. On a plane, points $A, B, C, D, E$ were marked such that triangles $A B C$ and $A D E$ are equal: $A B=A D, A C=A E, B C=D E$. Then, point $E$ and the labels of points $B, C$, and $D$ were erased from the drawing.

![](https://cdn.mathpix.com/cropped/2024_05_06_2763d6d141aaad15ef08g-04.jpg?height=603&width=606&top_left_y=90&top_left_x=428)

Let point $E$ have coordinates $(x ; y)$.

(a) (1 point) Indicate any possible value of the quantity $x y$.

(b) (3 points) Indicate all possible values of the quantity $x y$.
Answer: (a) any of the numbers $14,18,40$. (b) $14,18,40$.

Solution. From the condition $A B=A D$ it follows that $B$ and $D$ are the two rightmost marked points (in unknown order). The remaining unmarked point is $C$.

Case 1. Let point $B$ be at the top, and point $D$ at the bottom (Fig. 1a). It is clear that for point $E$ there are two options:

- point $E_{1}$ with coordinates $(8 ; 5)$
- point $E_{2}$ with coordinates $(6 ; 3)$.

Case 2. Let point $B$ be at the bottom, and point $D$ at the top. It is clear that for point $E$ there are two options:

- point $E_{3}$ with coordinates $(2 ; 7)$;
- point $E_{4}$ with coordinates $(6 ; 3)$.

It remains to note that points $E_{1}$ and $E_{4}$ coincide, so there are exactly three possible positions for point $E$, namely $(2 ; 7),(6 ; 3)$ and $(8 ; 5)$. Accordingly, the product of the coordinates of point $E$ can take exactly three values: 14,18 and 40.