Let $ABCD$ be a square. Choose points $E$ on $BC$ and $F$ on $CD$ so that $\angle EAF=45^\circ$ and so that neither $E$ nor $F$ is a vertex of the square. The lines $AE$ and $AF$ intersect the circumcircle of the square in the points $G$ and $H$ distinct from $A$, respectively. Show that the lines $EF$ and $GH$ are parallel.
Problem
Source: Bundeswettbewerb Mathematik 2019, Round 1 - Problem 3
Tags: geometry, circumcircle, square
sharp0407
06.08.2019 16:11
show that <AFE=<AHG (1) <AFD=<AFE(well-knowm, rotation) (2) tri OGH(45-45-90) --> <OHG=45 (3) let <OHA=a, <AHG=45+a, <FAD=45-a, <AFD=45+a (4) <AFE=<AFD=45+a=<AHG [end]
jayme
01.11.2019 11:08
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20III.pdf p. 8.. Sincerely Jean-Louis
dikhendzab
06.03.2021 20:25
Very beautiful problem! Let $\angle BAE=\angle CAH=\beta$ and $\angle BAC=\angle GAH=\angle CAD=\alpha=45^\circ$. From this it is obvious that $\angle GAC=\angle FAD=\alpha-\beta$ and because we have angles above the diameter $\angle AGC=\angle AHC=90^\circ$ and triangles $AGC$ and $AHC$ are right angled triangles. From them we can define some trigonometric values using previously defined angles: $AG=AC\cdot \cos(\alpha-\beta); AE=\frac{AB}{\cos \beta}; AH=AC\cdot \cos \beta; AF=\frac{AB}{\cos (\alpha-\beta)}$ From this we get: $\frac{AG}{AE}=\frac{AC\cdot \cos(\alpha-\beta)}{\frac{AB}{\cos \beta}}=\frac{AC \cdot \cos (\alpha-\beta)\cdot\cos \beta}{AB}=\frac{AH}{AF} \implies$ triangles $AEF$ and $AGH$ are similar and $\angle AEF=\angle AGH$ and $\angle AFE=\angle AHG \implies EF \parallel GH$