Let $EF$ cuts $AD $ and $BC$ at $K,K'$ , POP yields $\frac{KA_1}{KF}=\frac{KE}{KA}\overset{( KEA\sim K'EB)}{=} \frac{K'E}{K'B}$ and $\frac{K'C_1}{K'F}=\frac{K'E}{K'C}$ hence $\frac{KA_1}{K'C_1}=\frac{KF}{K'B}\cdot \frac{K'C}{K'F}$ but $\frac{K'C}{K'F}\overset{(K'CF\sim KDF)}{=}\frac{KD}{KF}$ thus $\frac{K'C_1}{KA_1}=\frac{KD}{K'B}$so if $A_1C_1,DB$ cut resp $EF$ at $X,Y$ then $\frac{K'X}{XK}=\frac{K'C_1}{KA_1}=\frac{KD}{K'B}=\frac{K'Y}{YK}$ therefore $X=Y$ which means that $BD,A_1C_1,EF$ are concurrent. RH HAS

proposed by Alexander Kuznetsov, Russia it was also RMM Shortlist 2017 G1

Romania TST 2017 P1 wrote: Let $ABCD$ be a trapezium, $AD\parallel BC$, and let $E,F$ be points on the sides$AB$ and $CD$, respectively. The circumcircle of $AEF$ meets $AD$ again at $A_1$, and the circumcircle of $CEF$ meets $BC$ again at $C_1$. Prove that $A_1C_1,BD,EF$ are concurrent. Solution: Rename $A_1,C_1$ as $A',C'$. Given, that, $AEFA'$ and $CFEC'$ are cyclic. Now WLOG assume, that points $A,D,A'$ and $C',B,C$ lie in the same order and assume, $DC > AB$. Let $\odot (C'BE)$ $\cap$ $\odot (AEFA')$ $=$ $K$, Hence, by Reim's Theorem $\implies$ $K$ $\in$ $A'C'$. Similarly, If $\odot (A'DF)$ $\cap$ $\odot (C'EFC)$ $=$ $N$ $\implies $By Reim's Theorem, $N$ $\in$ $A'C'$. Let $\odot (A'DF)$ $\cap$ $EF$ $=$ $G$ and let $G$ $\in$ $\overline{EF}$. Hence, By Reim's Theorem, $AE||DG$. Therefore, if $DG$ $\cap$ $BC$ $=$ $L$ $\implies$ $ABLD$ becomes parallelogram. Define: $BD$ $\cap$ $EF$ $=$ $M$. Let $BD$ $\cap$ $\odot (A'FD)$ $=$ $P$ $\implies$ $\angle DPF$ $=$ $\angle DGF$ $=$ $\angle AEF$ $\implies$ $BEPF$ is cyclic. Again By Reim's Theorem $\implies$ $C'BNP$ is cyclic. Hence, by Radical Axes Theorem on $\odot (C'BNP)$, $\odot (C'EFC)$ and $BEPF)$ $\implies$ $M$ lies on $A'C'$

Im new to this techniques so pls check my sketch someone.(high chances ive flawed).

n00b i am

Here's a completely different solution. 2017 RMMSL G1 wrote: Let $ABCD$ be a trapezium, $AD\parallel BC$, and let $E,F$ be points on the sides$AB$ and $CD$, respectively. The circumcircle of $AEF$ meets $AD$ again at $A_1$, and the circumcircle of $CEF$ meets $BC$ again at $C_1$. Prove that $A_1C_1,BD,EF$ are concurrent. For sake of Simplicity Rephrase the Problem as follows. Rephrased Problem wrote: Two Circles $\omega_1,\omega_2$ intersects at $E,F$ respectively. Let $\{A,A_1\}$ be any two points on $\omega_2$ and $\{C,C_1\}$ be any two points on $\omega_1$ such that $AA_1\|EF\|CC_1$ such that $\{A_1,A,E,C_1,C,F\}$ are in this order and if $CF\cap AA_1=D$ and $AE\cap CC_1=B$. Then $A_!C_1,BD,EF$ concurs. Key Claim:- $\{A_1,F,C,C_1,E,A\}$ lies on a unique conic $\mathcal C$. Proof:- Notice that $\{A_1F\cap AE,FC\cap EC_1,AC\cap A_1C_1\}$ all lie on the Perpendicular bisector of $EF$ as $CC_1EF,AA_1FE,AA_1CC_1$ are Isosceles Trapezoids. So, By Converse of Pascal's Theorem on $A_1FCAEC_1$ we get that $\{A_1,A,E,F,C,C_1\}$ lies on a unique conic $\mathcal C$. To Finish off apply Pascal on $AEFCC_1A_1$ to get that $A_1C_1,BD,EF$ are concurrent. $\blacksquare$

Here is a solution similar to my solution of IMO 2018 P1. Treat the case $EF\parallel BC$ by continuity. Let $X=EF\cap AD$ and $Y=EF\cap BC$. Notice that $$\measuredangle XA_1F = \measuredangle AA_1F = \measuredangle AEF = \measuredangle BEF$$therefore $\triangle XA_1F\sim\triangle YEB$, or $\tfrac{YE}{YB} = \tfrac{XA_1}{XF}$. Similarly, $\triangle XFD\sim\triangle YC_1E$, or $\tfrac{YE}{YC_1} = \tfrac{XD}{XF}$. Combining these two ratios gives $$XA_1\cdot YB = YE\cdot XF = XD\cdot YC_1 \implies \frac{XA_1}{XD} = \frac{YC_1}{YB},$$or $A_1C_1$, $BD$, $EF$ are concurrent.

Wrong moving points Solution @MarkBcc168 thanks

Nice idea, but I don't think that rephraised problem is the same. Original problem is more general: $EF$ is not parallel to $AA_1,CC_1$. But it can be easily fixed: Since $AA_1\parallel CC_1$ by Converse of Generalised Reim's Theorem, $AA_1CC_1EF$ lies on conic. So by Pascal on $AEFCC_1A_1$ we get $EF\cap A_1C_1\in BD$.

If I am not mistaken, both moving points solutions (by Geometrix and Functional_equation) are wrong. The first reason is both solutions used the case $EFDA$ and $EFBC$ being concyclic. This will result in $A_1=D$ and $C_1=B$. Thus, when $E$ approaches this point, the intersection of $A_1C_1$ and $BD$ can approach any point on $BD$, which we don't know whether it lies on $EF$ or not. Therefore, using this case is actually invalid. Furthermore, I want both solutions to elaborate on how to show that the map is projective. Both solutions begin with animating $E$ on $AB$. Then they proceed to show that $E\mapsto K_1 = EF\cap BD$ and $E\mapsto K_2 = A_1C_1\cap BD$ are projective. While I agree that the former map is obviously projective, I don't think the latter map is that obvious; both solutions claim that the sequence of projections are $E\mapsto AB\mapsto AD\mapsto A_1$. This is obviously wrong as $AB$ and $AD$ are both fixed lines. Furthermore, they claim that $A_1\mapsto K_2$ is projective. This is not immediately true as projection from $C_1$ is invalid, due to $C_1$ being moving points. Actually, it's true that $E\mapsto K_2$ is projective; the shortest proof I know is the following. Point $A_1$ has degree $1$, as $\measuredangle A_1FE$ is constant, therefore we can project from $F$ and rotate. Meanwhile, point $C_1$ has degree at least $2$. This is because, given a fixed position of $C_1$, one construct the working $E$ in two ways. Therefore this map is not bijective, hence can never be projective. Thus, $\overline{A_1C_1}$ has degree $3$, unless $A_1$ and $C_1$ are coincide at some point. However, this happens when $E_{\infty_{AB}}$; one can check that both $A_1, C_1$ are points at infinity. This means that $\overline{A_1C_1}$ actually has degree two. Since $\overline{A_1C_1}$ coincide with $\overline{BD}$ once, we get that $K_2$ has degree one. Thus, it suffices to find one more nontrivial case.

Let $EF\cap AD=X$ and $EF \cap BC=Y$, $BD \cap EF=K$, $A_1K \cap BC=C_1'$ Claim : $\triangle YEC_1'\sim \triangle XDF$ Proof : We have $\frac{YB}{XD}=\frac{BK}{DK}=\frac{BC_1'}{DA_1}$ So, $\frac{XF}{XD}=\frac{BC_1'}{DA_1}.\frac{XF}{YD}=\frac{BC_1'}{YB}.\frac{XA_1}{DA_1}.\frac{XA}{XE}$ (1) by power of point $X.$ $\triangle DXK \sim \triangle BYK \implies$ $\frac{XK}{YK}=\frac{DK}{BK}=\frac{DA_1}{BC_1'}$ and $\triangle A_1XK \sim C_1'YK \implies \frac{XK}{YK}=\frac{XA_1}{YC_1'}$, so we have $YC_1'=\frac{XA_1}{DA_1}.BC_1'$ (2) $\triangle YEB \sim \triangle XEA \implies$ $YE=\frac{XE}{XA}.YB$ (3) From (2) and (3), we have $\frac{YC_1'}{YE}=\frac{XA_1}{DA_1}.\frac{BC_1'}{YB}.\frac{XA}{XE}=\frac{XF}{XD}$ from (1) So, $\triangle YEC_1'\sim \triangle XDF$ Therefore $\angle YC_1'E =\angle DFX =\angle EFC$ or points $C,C_1',E,F$ are concyclic and we are done.

Let $P = \overline{AB} \cap \overline{CD}$. By Menelaus it suffices to show that $$\frac{BC_1}{A_1D} = \frac{PF}{FD} \cdot \frac{BE}{PE}.$$This is equivalent to \begin{align*} 1 &= \frac{A_1D}{FD} \cdot \frac{BE}{BC_1} \cdot \frac{PF}{PE} \\ &= \frac{\sin \angle A_1FD}{\sin \angle PEF} \cdot \frac{\sin \angle PFE}{\sin \angle C_1EB} \cdot \frac{\sin \angle PEF}{\sin \angle PFE} \\ &= \frac{\sin \angle A_1FD}{\sin \angle C_1EB}. \end{align*}Thus, it suffices to show that $$180^\circ - \angle A_1FD = \angle EAD + \angle EFD = \angle BEF+ \angle BCF = 180^\circ - \angle C_1EB.$$This follows because $$\angle EAD - \angle BCF = \angle EAD - \angle PDA = \angle P = \angle BEF - \angle EFD,$$as required.