It is $CGMO$ 2007/5 Assume that $AF=2$ and $FC=1$ so los on $\triangle ADC$ gives $AD=DC=\sqrt 3$ and let $O$ be the circumcentre of $\triangle ADC$ then $\angle AOD=120$ and los on $\triangle AOD$ gives $OD=AO=1$ and $OD \parallel AC$ so $ODFC$ is a parallelogram. Let OC intersects $AD$ at $X$ then $DX=XF$ . Now we need to prof that $DX=EX$ computation gives $XD=XF=1/2$ and in $\triangle OCB$,$XE=1/2BO=1/2$ so $XD=DF=EF$ so proved

Take homothety at $C$ with $k=2$, mapping $E \mapsto B, D \mapsto K, F \mapsto G$. So now we need to prove that $\angle KBG = 90^\circ$. Very easy observations gives: $\triangle ADK$ - equilateral, $\angle CAK = 90^\circ$, $DBKA$ - cyclic. If $A$ midpoint of segment $KK'$, you can see that $G$ - centroid of equilateral $\triangle CKK'$ ($CG=2AG$) and so also its incenter, $\implies \angle AGK = 60^\circ$ , $G \in (DBKA)$ , $\angle KBG = 90^\circ$.

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zuss77 wrote: Take homothety at $C$ with $k=2$, mapping $E \mapsto B, D \mapsto K, F \mapsto G$. So now we need to prove that $\angle KBG = 90^\circ$. Very easy observations gives: $\triangle ADK$ - equilateral, $\angle CAK = 90^\circ$, $DBKA$ - cyclic. If $A$ midpoint of segment $KK'$, you can see that $G$ - centroid of equilateral $\triangle CKK'$ ($CG=2AG$) and so also its incenter, $\implies \angle AGK = 60^\circ$ , $G \in (DBKA)$ , $\angle KBG = 90^\circ$. What is $\text{homothety}$?

JustKeepRunning wrote: What is $\text{homothety}$? You are kidding, right?

zuss77 wrote: JustKeepRunning wrote: What is $\text{homothety}$? You are kidding, right? No...

Homothety is a brilliant geometric transformation that is very useful to solving difficult problems. Definition: Let $O$ a stable point in plane and a number $k>0$. Also let $A$ a random point of the plane. In $OA$ we take $A'$ such that $OA'=kOA$. The transformation (image) such that in every point $A$ there is a point $A'$ such that $OA'=kOA$ is called homothety with centre $O$ and ratio $k$. https://en.wikipedia.org/wiki/Homothetic_transformation https://brilliant.org/wiki/euclidean-geometry-homothety/ These will help you in the beginning.

minageus wrote: Homothety is a brilliant geometric transformation that is very useful to solving difficult problems. Definition: Let $O$ a stable point in plane and a number $k>0$. Also let $A$ a random point of the plane. In $OA$ we take $A'$ such that $OA'=kOA$. The transformation (image) such that in every point $A$ there is a point $A'$ such that $OA'=kOA$ is called homothety with centre $O$ and ratio $k$. https://en.wikipedia.org/wiki/Homothetic_transformation https://brilliant.org/wiki/euclidean-geometry-homothety/ These will help you in the beginning. Oh ok thanks

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