Let $ABC$ be a isosceles triangle with $\overline{AB}=\overline{AC}$. Let $D(\neq A, C)$ be a point on the side $AC$, and circle $\Omega$ is tangent to $BD$ at point $E$, and $AC$ at point $C$. Denote by $F(\neq E)$ the intersection of the line $AE$ and the circle $\Omega$, and $G(\neq a)$ the intersection of the line $AC$ and the circumcircle of the triangle $ABF$. Prove that points $D, E, F,$ and $G$ are concyclic.
Problem
Source: KJMO 2022 P6
Tags: geometry, circumcircle, Concyclic, tangent
keglesnit
29.10.2022 20:16
sol $AE\cdot AF=AC^2=AB^2$ implies $AB$ tangent to $(BEF)$ at $B$. This gives $\angle BGA=\angle BFA=\angle BFE=\angle ABE=\angle ABD$ which yields $AB$ tangent to $(BDG)$ at $B$. Now we are done since $AD\cdot AG=AB^2=AE\cdot AF$
john0512
20.02.2023 22:10
huh what lol Let $\angle AEB=\alpha$. By Power of a Point, $AE\cdot AF=AC^2=AB^2$. Since $AE\cdot AF=AB^2$, $\triangle ABE\sim\triangle AFB.$ Therefore, $$\angle ABF=\angle AEB=\alpha.$$From $ABFG$ cyclic, $\angle AGF=180-\alpha$, and we also know that $\angle DEF=\alpha$, so we are done.
lian_the_noob12
21.04.2023 13:11
john0512 wrote: huh what lol Let $\angle AEB=\alpha$. By Power of a Point, $AE\cdot AF=AC^2=AB^2$. Since $AE\cdot AF=AB^2$, $\triangle ABE\sim\triangle AFB.$ Therefore, $$\angle ABF=\angle AEB=\alpha.$$From $ABFG$ cyclic, $\angle AGF=180-\alpha$, and we also know that $\angle DEF=\alpha$, so we are done. Joss -,-