3. Inside side $BC$ of the convex quadrilateral $ABCD$, there is a point $E$ such that the line $AE$ divides the quadrilateral into two equal area parts. Which vertex of the quadrilateral is farthest from the line $AE$? (I. Rubanov, D. Shiryayev)
Answer. Vertex $B$. Solution. Drop perpendiculars $B B_{1}, C C_{1}, D D_{1}$ to the line $A E$. Then

$$
D D_{1} \cdot A E=2 S(A D E)<2 S(A D C E)=2 S(A B E)=B B_{1} \cdot A E
$$

from which $D D_{1}<B B_{1}$. Similarly, from the inequality $2 S(A C E)<2 S(A D C E)$, we get that $C C_{1}<B B_{1}$. Since the distance from point $A$ to the line $A E$ is 0, we get that

![](https://cdn.mathpix.com/cropped/2024_05_06_33d376b8c98a80a4571cg-1.jpg?height=257&width=351&top_left_y=1232&top_left_x=1593)
vertex $B$ is farther from the line $A E$ than all other vertices of the quadrilateral $A B C D$.