Let $ABCDEF$ be a convex hexagon such that $\angle ACE = \angle BDF$ and $\angle BCA = \angle EDF$. Let $A_1=AC\cap FB$, $B_1=BD\cap AC$, $C_1=CE\cap BD$, $D_1=DF\cap CE$, $E_1=EA\cap DF$, and $F_1=FB\cap EA$. Suppose $B_1, C_1, D_1, F_1$ lie on the same circle $\Gamma$. The circumcircles of $\triangle BB_1F_1$ and $ED_1F_1$ meet at $F_1$ and $P$. The line $F_1P$ meets $\Gamma$ again at $Q$. Prove that $B_1D_1$ and $QC_1$ are parrallel. (Here, we use $l_1\cap l_2$ to denote the intersection point of lines $l_1$ and $l_2$)
Problem
Source:
Tags: geometry, circumcircle
Britishgoat
12.12.2016 16:01
How is $D_1$ defined?
YanYau
12.12.2016 16:14
sorry about that, $D_1$'s definition is added on now.
Britishgoat
12.12.2016 21:02
By angles in circumferences $\angle C_1EP=\angle D_1F_1P$ and $\angle PBC_1=\angle PF_1B_1$ and $\angle D_1F_1B_1+\angle D_1C_1B_1=180$ so in quadrilateral $EPBC_1$: $$\angle PEC_1+\angle EC_1B+\angle C_1BP+\angle BPE=360$$and it results that $\angle BPE=180$ It can be easily seen that $\triangle EC_1D$ and $\triangle BC_1C$ are similar. Also, because of the angles given $B_1$ and $D_1$ are homologous points, so: $$\frac{D_1C_1}{ED_1}=\frac{C_1B_1}{B_1B}$$Which means that $B_1D_1\parallel EB$. Then $\angle D_1F_1C_1=\angle D_1B_1C_1=\angle PBB_1=\angle P_1F_1Q$. Finally $\stackrel{\frown}{D_1C_1} =\stackrel{\frown}{QB_1} $ so $C_1Q\parallel B_1D_1$.
Tumon2001
03.04.2017 21:17
Solution: Trivial angle chasing proves that $C_1Q || EP $. Also, $\angle EPF_1 + \angle BPF_1 = \angle ED_1F_1 + \angle BB_1F_1 = 180^{\circ} $. So, $B $, $P $ and $E $ are collinear. Again, observe that $CB_1D_1D $ and $CBED$ are cyclic. So, $B_1D_1 || BE $. But, $BE || C_1Q $. Hence, $B_1D_1 || C_1Q $.
mathaddiction
18.08.2021 04:52
[asy][asy] size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.92, xmax = 8.92, ymin = -1.58, ymax = 8.9; /* image dimensions */ /* draw figures */ draw((-0.94,1.06)--(-6.4,2.42), linewidth(0.8)); draw((-0.94,1.06)--(0.46,3.64), linewidth(0.8)); draw((-6.4,2.42)--(0.46,3.64), linewidth(0.8)); draw(circle((-2.460180359790245,3.045215386214736), 2.5004056582794165), linewidth(0.8)); draw(circle((-5.14482480745339,3.907304588978874), 1.9461602463266063), linewidth(0.8)); draw(circle((-1.674138209033565,4.288525847044444), 2.2305003182115235), linewidth(0.8)); draw((-3.2026849736045415,0.6575982991548771)--(-3.7032424817307152,5.214737226840587), linewidth(0.8)); /* dots and labels */ dot((-0.94,1.06),dotstyle); label("$C_1$", (-0.86,1.26), NE * labelscalefactor); dot((-6.4,2.42),dotstyle); label("$B$", (-6.32,2.62), NE * labelscalefactor); dot((0.46,3.64),dotstyle); label("$E$", (0.54,3.84), NE * labelscalefactor); dot((-4.733944903275871,2.005011917299484),dotstyle); label("$B_1$", (-4.66,2.2), NE * labelscalefactor); dot((0.032806385455351625,2.8527431960534333),linewidth(4pt) + dotstyle); label("$D_1$", (0.12,3.02), NE * labelscalefactor); dot((-3.7032424817307152,5.214737226840587),dotstyle); label("$F_1$", (-3.62,5.42), NE * labelscalefactor); dot((-3.4538192280939097,2.943956347190294),linewidth(4pt) + dotstyle); label("$P$", (-3.38,3.1), NE * labelscalefactor); dot((-3.2026849736045415,0.6575982991548771),linewidth(4pt) + dotstyle); label("$Q$", (-3.12,0.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Notice that $$\angle BPF_1+\angle EPF_1=\angle BB_1F_1+\angle ED_1F_1=180^{\circ}$$By Reim's theorem $B_1D_1\|BE$, therefore $$\angle B_1F_1P=\angle B_1BP=\angle C_1B_1D_1=\angle C_1F_1D_1$$as desired.
HamstPan38825
03.09.2022 03:59
Despite looking scary, this problem is quite simple. The given condition implies that $(BCDE)$ and $(B_1D_1DC)$ are concyclic. The main claim is that $B, P, E$ are collinear. This is because $$\measuredangle F_1PB = \measuredangle F_1B_1B = \measuredangle F_1B_1D = \measuredangle F_1D_1C_1 = \measuredangle F_1D_1E = \measuredangle F_1PE,$$yay. So then $$\measuredangle BF_1Q = \measuredangle BF_1P = \measuredangle B_1BE = \measuredangle DCE = \measuredangle DB_1D_1 = \measuredangle C_1FD_1$$by the claim, so we are done.