Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National olympiad Prove it Mongolia

Diagonals of the convex quadrilateral $ABCD$ inscribed in a circle intersect at point $E$ . The line passing $E$ perpendicular to the side $AD$ intersects the side $BC$ at point $N$ . The line passing $E$ perpendicular to the side $BC$ intersects the side $AD$ at point $M$ . Prove that $2MN = AB + CD$ implies that $ABCD$ is a trapezoid.

(proposed by N. Argilsan)

Solution

Let denote $\angle BAD = \alpha$ . Then $\angle BCA = BAD = \alpha$ .

Since $\angle BCA = 90^\circ - \alpha = \angle AEM \Rightarrow \angle MED = \alpha$ and we conclude that $\triangle EMD$ is. Similarly, it is easy to show $AM = EM = MD$ . In other words, point $M$ is the midpoint of $AD$ . Similarly, $N$ is the midpoint of $BC$ . Denote $P$ as the midpoint of $AC$ . It implies that $NP = \frac{1}{2}AB$ and $MP = \frac{1}{2}CD$ .

Therefore the inequality $MN \le NP + MP = \frac{1}{2}(CD + AB)$ holds. If points $N$ , $P$ , $M$ lie on a line then the $MN = \frac{1}{2}(CD + AB)$ equality holds and $NP \parallel MP \Rightarrow NP \parallel AB$ and $MP \parallel CD \Rightarrow CD \parallel AB$ . It means $ABCD$ is a trapezoid.

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