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Geometry Difficulty 5.2 AIME, harder Prove it Vietnam

Let be given a convex quadrilateral ABCD. A point M moves on the line AB but does not coincide with A and B. Let N be the second point of meeting (distinct from M) of the circles (MAC) and (MBD). Prove that

i) $N$ moves on a fixed circle,

ii) The line $MN$ passes through a fixed point.

(The symbol $(XYZ)$ denotes the circle passing through the points $X$ , $Y$ , $Z$ .)

Solution

Let $I$ be the point of intersection of the two diagonals of the quadrilateral $ABCD$ (figure 1).

i) The quadrilateral $DCIN$ is cyclic because $\widehat{ICN} = \widehat{IDN}$ (as both angles are equal to $\widehat{AMN}$ ). Consequently, $N$ moves on the fixed circle.

ii) Draw the line $l$ passing through $I$ parallel to $AB$ . It cuts $MN$ at $K$ . The quadrilateral $IKCN$ is cyclic because $\widehat{ICN} = \widehat{IKN}$ (as both angles are equal to $\widehat{AMN}$ ). Therefore, the second meeting point of $l$ with the circle $(IDC)$ passes through a fixed point.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.