Let $\omega$ is tangent to the sides of an acute angle with vertex $A$ at points $B$ and $C$. Let $D$ be an arbitrary point onn the major arc $BC$ of the circle $\omega$. Points $E$ and $F$ are chosen inside the angle $DAC$ so that quadrilaterals $ABDF$ and $ACED$ are inscribed and the points $A,E,F$ lie on the same straight line. Prove that lines $BE$ and $CF$ intersectat $\omega$.
Problem
Source: VII Caucasus Olympiad, Senior, Day1 P4.
Tags: geometry
Jalil_Huseynov
13.03.2022 22:37
Again easy P4 (Like lsast year's P4). But anyway, it's better than last year's P4. Invert around circle $(A,AB)$ and assume $X'$ is image of $X$ under this inversion. We get $D'=AD\cap \omega, E'=AE\cap CD', F'=AF\cap BD'$ and let $T=BE\cap CF$. Since $AE\cdot AE'=AC^2$ we get $(CEE')$ tangents to $AC \implies \angle E'EC=\angle E'CA=\angle D'BC \implies BCEF'$ is cyclic. Similarly $BCFE'$ is cyclic. So $\angle TEF=\angle BEF'=\angle BCF'$ and $\angle F'FC=\angle F'CA \implies \angle BTC=\angle TEF + \angle EFT = \angle BCF'+\angle F'CA=\angle BCA \implies T\in \omega$. So we are done!
VicKmath7
13.03.2022 22:37
Well, almost trivial. Note that there is a spiral similarity taking $DEF$ to $DBC$ (more formally, we have that triangles $DBE$ and $DCF$ are also similar, so if $BE$ intersects $CF$ at $P$, then $BCDP$ is cyclic).
Mahdi_Mashayekhi
02.04.2022 13:41
Easy for P4... $\angle DBC = \angle DCF = \angle 180 - \angle ACD = \angle 180 - \angle AED = \angle DEF$ and $\angle BCD = \angle 180 - \angle ABD = \angle EFD$ so $DBC$ and $DEF$ have spiral similarity. Let $BE$ meet $CF$ at $X$. It's well known that $DEXF$ and $DBCX$ are cyclic so $X$ lies on $\omega$.
Assassino9931
19.05.2024 18:32
Equivalent, but without the use of spiral similarity basic trickery (I know it is easier to figure out with it, but just for the sake of inexperienced people). We have $\angle DFE = 180^{\circ} - \angle AFD = \angle ABD = \angle BCD$ from the cyclic $ABDF$ and the tangency around $\omega$, similarly $\angle AED = \angle ACD = \angle DBC$. Hence $\triangle DBC \sim \triangle DEF$. Hence $\frac{BD}{DC} = \frac{DE}{DF}$, i.e. $\frac{BD}{DE} = \frac{DC}{DF}$ and also $\angle BDC = \angle EDF$, i.e. $\angle BDE = \angle CDF$. Hence $\triangle BDE \sim \triangle CDF$, so $\angle DBE = \angle DCF$, thus after extending $BE$ and $CF$ to intersect $\omega$, the respective arcs would be equal and so the intersection points would coincide, done.
Assassino9931
02.06.2024 02:54
Plot twist: only angle chasing suffices! (Credits to @africanboy) We have $\angle DFE = 180^{\circ} - \angle AFD = \angle ABD = \angle BCD$ from the cyclic $ABDF$ and the tangency around $\omega$, similarly $\angle AED = \angle ACD = \angle DBC$ and hence $\angle BDC = \angle EDF$. Now define $T = BE \cap \omega$. Then $\angle DTE = \angle BTD = \angle BCD = \angle DFE$, hence $DFTE$ is cyclic. From here $\angle ETF + \angle CTE = (180^{\circ} - \angle FDE) + \angle BTC = (180^{\circ} - \angle BDC) + \angle BTC = 180^{\circ}$, so $CE$ passes through $T$ and we are done.