$w_1$ and $w_2$ circles have different diameters and externally tangent to each other at $X$. Points $A$ and $B$ are on $w_1$, points $C$ and $D$ are on $w_2$ such that $AC$ and $BD$ are common tangent lines of these two circles. $CX$ intersects $AB$ at $E$ and $w_1$ at $F$ second time. $(EFB)$ intersects $AF$ at $G$ second time. If $AX \cap CD =H$, show that points $E, G, H$ are collinear.
Problem
Source: 2021 Turkey JBMO TST P8
Tags: geometry, geometry proposed
hakN
24.05.2021 10:10
Let $J = AX\cap w_2 \neq X$.
Using the homothety and a bit of angle chasing, one gets $\angle AXC = \angle AXF = 90$. Also angle chasing shows that $AEJC$ is cyclic. So $JE \perp AE$ and since $CD\parallel AE$ and $JD\perp CD$, we get $JD\perp AE$ and so $J,D,E$ are collinear.
So now $90 = \angle EXH = \angle HDE$ we have $EXHD$ is cyclic.
So we have $\angle XEH = \angle XDH = \angle XDC = \angle XJC = 90 - \angle XAC = \angle FBG = \angle FEG = \angle XEG$ and so $E,G,H$ are collinear as desired. $\square$
Attachments:
aleksijt
31.05.2021 12:29
First, $\angle CXA=90$, so $\angle FBA=90$ and $\angle FGE=90$. Now, $AEGX$ is cyclic. Note that if $H$, $F$ and $B$ are collinear, then by looking at the radical center of $\omega_{1}$, $(FBEG)$ and $(AEGX)$ we'll get the collinearity wanted. Claim: $H$, $F$ and $B$ are collinear. For this, we will use that $\triangle HXD \sim \triangle FXB$. (It follows directly by some angle chasing.) $XF : XH = XB : XD$ and $\angle BXD = \angle FXH$ so $\triangle BXD \sim \triangle FXH$ $\implies$ $\angle XFH = \angle XBD = \angle XAB = 180 - \angle XFB$.
SerdarBozdag
31.05.2021 16:23
Let $O_1$ be the center of $\omega_1$. Because $X$ is the midpoint of arc $CD$ we have $\angle ACX=\angle XCH$ and $\angle AXC=90 \implies AX=XH$, $EX=XC \implies ACHE$ is a rhombus. $\angle AXC=90 \implies O_1$ is on $A-F-G$. $\angle EBF=90 \implies \angle EGF=90 \implies AXGE$ is cyclic. $$\angle XEH=\angle XCA=\angle XAF=\angle FEG=\angle XEG$$Thus $E-G-H$ are collinear.