Let $ ABC$ a triangle. Let $D$ on $AB$ and $E$ on $AC$ such that $DE||BC$. Let line $DE$ intersect circumcircle of $ABC$ at two distinct points $F$ and $G$ so that line segments $BF$ and $CG$ intersect at P. Let circumcircle of $GDP$ and $FEP$ intersect again at $Q$. Prove that $A, P, Q$ are collinear.
Problem
Source: Junior Olympiad of Malaysia Shortlist 2015 G3
Tags: geometry
gavrilos
17.07.2015 17:37
Hello! Suppose that $G,D,E,F$ are in this order (if they are in the order $F,D,E,G$ the proof is the same). Since $PQ$ is the radical axis of the circles $c_{1}:(G,D,P)$ and $c_{2}:(F,E,P)$, it suffices to show that $A$ lies in the radical axis of these circles. First,notice that $BCFG$ is an isosceles trapezoid,thus $\angle{PGF}=\angle{PFG}=\angle{PBC}=\angle{PCB}$. Suppose that the line $AB$ cuts again $c_{1}$ at $K$ and that the line $AC$ cuts again $c_{2}$ at $L$. We have $\angle{PKB}=\angle{PKD}=\angle{PGD}=\angle{PCB}$ thus $KBPC$ is cyclic thus $\angle{PKC}=\angle{PBC}=\angle{PFE}=\angle{PLE}$. The latter equality gives that $PKCL$ is cyclic,thus $\angle{PKL}=\angle{PCL}$. Hence $\angle{LKD}=\angle{LKB}=\angle{PKB}+\angle{PKL}=\angle{PCB}+\angle{PCL}=\angle{BCL}\overset{DE\parallel BC}=\angle{AED}$. Thus $DELK$ is cyclic,which gives that $AD\cdot AK=AE\cdot AL$,that is,$A$ has equal powers with respect to $c_{1},c_{2}$. This implies that it belongs to their radical axis,which is the desired result.
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Jzhang21
10.08.2018 05:12
This is a nice problem. Proof: Let $\omega_1 =(FEP)$ and $\omega_2=(GDP). $ It suffices to prove that $A$ lies on the radical axis of $\omega_1$ and $\omega_2$. Let the second intersection of $AD ,AE$ and $\omega_2,\omega_1$ be $D'$ and $E'$ respectively. By Radical Lemma, it suffices to prove that $DD'EE'$ is cyclic. Since $\measuredangle BD'P=\measuredangle DD'P=\measuredangle DGP=\measuredangle BCP, $ $BD'PC $is cyclic. Since $\measuredangle E'CP=\measuredangle E'EF=\measuredangle E'PB=\measuredangle E'PB,$ $BE'PC$ is cyclic. As a result, $\measuredangle BD'C=\measuredangle BPC=\measuredangle BE'C$ so $BD'CE'$ is cyclic. Thus, $\measuredangle BD'E'=\measuredangle BCE'=\measuredangle DEE'$ so $DD'EE'$ is cyclic, as desired. $\blacksquare$
Mahdi_Mashayekhi
10.02.2022 09:35
Note that PQ is radical axis. Let AB meet GPD at S and AC meet FPE at K. we will prove DEKS is cyclic. Claim1 : SBCKP is cyclic. Proof : ∠BSP = ∠DSP and ∠BCP = ∠DGP and ∠DSP + ∠DGP = 180 so BCPS is cyclic. with same approach we can prove BCKP is cyclic. ∠DSK = ∠BSK = ∠BPK and ∠DEK = ∠BCK and ∠BPK + ∠BCK = 180 so DEKS is cyclic as wanted.