Tangent and circles with equal angles

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Tangent and circles with equal angles

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Source: 2024 CWMO P3

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#1 h Aug 6, 2024, 4:52 AM • 2 Y

Y by Rounak_iitr, ehuseyinyigit

$AB,AC$ are tangent to $\Omega$ at $B$ and $C$ , respectively. $D,E,F$ lie on segments $BC,CA,AB$ such that $AF<AE$ and $\angle FDB= \angle EDC$ . The circumcircle of $\triangle FEC$ intersects $\Omega$ at $G$ and $C$ . Show that $\angle AEF= \angle BGD$

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egxa

#2 h Aug 6, 2024, 12:44 PM • 5 Y

Y by GioOrnikapa, iamnotgentle, sami1618, ehuseyinyigit, Rounak_iitr

solved with binsherlo
Let $CG\cap EF =L, EF\cap BC=K$ and $N$ be the reflection of $K$ to $L$ .
$\angle BGF=\angle BGC-\angle FGC=\angle BCA-\angle KEC=\angle BKF$ Hence $B,F,G,K$ are cyclic.It's easy to show that $\angle CKL=\angle LGK$ . We have $LN^2=LK^2=LC.LG=LE.LF$ thus, $(K,N;E,F)=-1$ .
$BDF\sim CDE$ and $FNE\sim BDC$ .
$\frac{DE}{DF}=\frac{CD}{BD}=\frac{EN}{FN}$ Which gives that $\angle FDN=\angle NDE$ . So $ND\perp BC\implies LD=LN=LK$ .
$\angle LDK=\angle LGK$ so $L,D,G,K$ are cyclic. $\angle DGL=\angle DKL=\angle BGH\implies \angle AEF=\angle BGD$ .

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#3 h Aug 11, 2024, 3:36 PM • 1 Y

Y by ayeen_izady

Let $X$ be the intersection of $CG$ and $EF$ , let $Y$ be the intersection of $BC$ and $EF$ , and let $D'$ be the fixed point along segment $BC$ of the inversion centered at $X$ preserving $(FECG)$ .

Now $\angle GBD'=180^{\circ}-\angle BCG-\angle BGC =\angle BCX-\angle BCA=\angle XCE=\angle GFX$ $\angle GBF=\angle ABC+\angle GBD'=\angle ACB +\angle XCE=\angle XCD'=\angle GD'X$ These properties uniquely define $D'$ so $D'$ must be the image of $F$ in $\sqrt{xb}$ -inversion in triangle $GXB$ . As $\angle ED'C=\angle XFD'+\angle XGD'=\angle EFD'+\angle FGB=\angle EFD'+\angle BGC-\angle CGF=$ $\angle EFD'+\angle ACB-\angle AEF=\angle EFD'+\angle FYB=\angle FD'B$ Thus we have that $D=D'$ to finish the problem simply notice $\angle AEF=\angle XGF=\angle BGD$

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#4 h Oct 3, 2024, 2:12 AM

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Let $P$ and $T$ be the intersections of $EF$ with $CG$ and $BC$ , respectively. Note that
$\angle GBT=180^\circ-\angle ECG=\angle GFT,$ so $BGTF$ is cyclic. From this, $\angle TGF=\angle TBF=\angle CGB$ , so $\triangle BGC\sim\triangle FGT$ , and $\angle PTC=\angle FGB=\angle TGC,$ implying that $PT^2=PC\cdot PG=PE\cdot PF.$ After some manipulations, this is equivalent to $\frac{FT}{TE}=\frac{FP}{PT}$ . By Menelaus on $\triangle AFE$ with the colinear points $B-C-T$ and the similar triangles $\triangle FBD\sim\triangle ECD$ , this is equivalent to $\frac{FP}{PT}=\frac{BD}{DC}$ . Hence, $D$ and $P$ are corresponding points in the similar triangles $BGC$ and $FGT$ , so the result follows. $\square$

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#5 h Nov 18, 2024, 2:03 PM • 1 Y

Y by Rounak_iitr

Let $FE$ meet $BC$ at $P$ , $FG$ meet $\Omega$ again at $Q$ . Then we have
$\angle AEF=\angle CGF=\angle ECQ,$ i.e. $EF\parallel CQ$ . This implies $\angle FPB=\angle QCB=\angle FGB$ , i.e. $F,B,G,P$ are concyclic. Hence,
$\angle CGP=\angle BCG-\angle BPG=\angle BQG-\angle BFG=\angle FBQ=\angle BCQ=\angle FPB,$ i.e. $FP$ touches $(PCG)$ .

Let $X$ be a point on line $EF$ such that $DX=XP$ . Then by angle chasing we have
$\angle EDX=\angle EDC-\angle XDP=\angle FDB-\angle XPD=\angle DFE,$ i.e. $DX$ touches $(DEF)$ .

This implies $X$ lies on the radical axis $\ell$ of $(PCG)$ and $(DEF)$ .
Note that $C,E,F,G$ are also concyclic, implying $EF, CG$ and $\ell$ must concur at $X$ .

Hence, $DX^2=PX^2=XC\cdot XG$ , implying $\angle DGX=\angle XDC=\angle XPD=\angle QGB$ . So we obtain that $\angle BGD=\angle QGC=\angle AEF$ as desired.

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