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Taiwan IMO Selection Camp

Geometry Difficulty 5.2 AIME, harder Prove it Taiwan

Let quadrilateral $ABCD$ have an inscribed circle with center $I$ . Let the diagonals $AC$ , $BD$ intersect at point $E$ .
If the midpoints of the three segments $AD$ , $BC$ , $EI$ are collinear, prove that $AB = CD$ .

Solution

Let $AB$ and $CD$ intersect at point $F$ , and let the midpoints of segments $EF$ , $AD$ , $BC$ be $X$ , $Y$ , $Z$ respectively.
By Gauss's theorem, the three points $X$ , $Y$ , $Z$ are collinear (the Newton line). The line originally containing $X$ , $Y$ , $Z$ passes through point $F$ ; applying a scaling by factor $2$ centered at point $E$ will make this line pass through point $I$ . Therefore $YZ \parallel FI$ .
Let the midpoints of $AC$ , $BD$ be $U$ , $V$ respectively. Since $FI$ is the angle bisector of $\angle DFA$ , $YZ$ is also the angle bisector of $\angle UYV$ . But $YZ$ bisects segment $UV$ ( $UYVZ$ is a parallelogram), so $UYVZ$ is a rhombus, which gives $YU = YV$ . Therefore $AB = 2YU = 2YV = CD$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.