Library / Croatia Mathematical Competitions /18 of 25

Croatia Mathematical Competitions

Geometry Difficulty 6.0 National Olympiad Prove it Croatia

On the side $AC$ of the triangle $ABC$ the points $D$ and $E$ are given such that $D$ is between $C$ and $E$ . Let $F$ be the intersection of the circumcircle of the triangle $ABD$ and the line through the point $E$ parallel to $BC$ such that $E$ and $F$ are on different sides of the line $AB$ . Let $G$ be the intersection of the circumcircle of the triangle $BCD$ and the line through $E$ parallel to $AB$ such that $E$ and $G$ are on different sides of the line $BC$ .
Prove that the points $D, E, F$ and $G$ lie on the same circle.

Solution

Let $F'$ be the intersection of the circumcircle of the triangle $ABD$ and the line $BG$ (different from $B$ ).

The quadrilateral $DAF'B$ is cyclic, so we have $\angle BF'D = \angle BAD = \angle BAC$ . Since $GE \parallel AB$ , we have $\angle BAC = \angle GEC$ . Hence $\angle GF'D = \angle GEC$ , which means that $DEF'G$ is a cyclic quadrilateral.
Therefrom $\angle AEF' = \angle DGF' = \angle DGB$ . Since the quadrilateral $CDBG$ is cyclic, we have $\angle DGB = \angle DCB$ , so we can conclude $F'E \parallel BC$ .
Hence, $F'$ is the intersection point of the circumcircle of the triangle $ABD$ and the line parallel to $BC$ through $E$ , which means that $F' = F$ . Hence $DEFG$ is a cyclic quadrilateral, which means that $D, E, F$ and $G$ lie on the same circle.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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