Let $ABC$ be an acute triangle with $AB<AC < BC$, inscirbed in circle $\Gamma_1$, with center $O$. Circle $\Gamma_2$, with center point $A$ and radius $AC$ intersects $BC$ at point $D$ and the circle $\Gamma_1$ at point $E$. Line $AD$ intersects circle $\Gamma_1$ at point $F$. The circumscribed circle $\Gamma_3$ of triangle $DEF$, intersects $BC$ at point $G$. Prove that: a) Point $B$ is the center of circle $\Gamma_3$ b) Circumscribed circle of triangle $CEG$ is tangent to $AC$.
Problem
Source: 2022 Greece JBMO TST p2
Tags: geometry, tangent
Lemmas
05.05.2023 08:49
Just angle chasing
Batapan
05.05.2023 16:44
a) By the cyclic $AFBC$ and since $DA=AC$ , $\angle C=\angle BDF=\angle BFD$ so $BD=BF$ Also, $\angle DEB=\angle EBG-\angle EDB=\angle EBC- \frac{\angle EAC}{2}=\frac{\angle EAC}{2}=\angle EDB$ so $DB=BE$ Thus $B$ is the circumcenter of $\triangle DEF$ b) Consider the extension of $AC$, $ACx$ We have $\angle EGC=180-\angle DGE=180- \angle DFE$ and $\angle DFE=\angle ACE=180-\angle ECx$ $\implies \angle EGC=\angle ECx$ which proves the tangency.
Attachments:
Alex009
27.08.2024 21:39
a) \[\angle FDB=\angle ADB=\angle ADC=\angle ACD=\angle ACB=\angle DFB\]$\implies BD=BF$ \[\angle EBC=\angle EAC=\angle 2EDC\]$\implies BE=BD=BF$ $\blacksquare$ b) \[\angle GEC=\angle DGE -\angle DCE=\angle DFE -\angle DCE=\angle ACE -\angle DCE=\angle DCA=\angle BCA\]$\blacksquare$