Let $ABCD$ be a cyclic quadrilateral with $AB=BC$ and $AD = CD$. A point $M$ lies on the minor arc $CD$ of its circumcircle. The lines $BM$ and $CD$ meet at point $P$, the lines $AM$ and $BD$ meet at point $Q$. Prove that $PQ \parallel AC$.
Problem
Source: Sharygin Finals 2017, Problem 8.1
Tags: geometry, Projective
04.08.2017 19:34
Let $N=MA \cap CD$ and $PQ \cap AC=Z$; project $$-1=(AC, BD) \overset{M}{=} (NC, QD) \overset{P}{=} (AC, ZL),$$where $L$ is the midpoint of $AC$. Hence, $PQ \parallel AC$; just as we wished. $\blacksquare$
06.08.2017 08:38
1line solution by Reim's Theorem.
06.08.2017 08:41
tarzanjunior wrote: 1line solution by Reim's Theorem. What's that?
06.08.2017 16:53
$\measuredangle QMP = \measuredangle AMB = \measuredangle ACB = \measuredangle BAC = \measuredangle BDC = \measuredangle QDP$. So, $M, D, Q, P$ are concyclic. Also, $M, D, A, C$ are cyclic. So, by Reim's Theorem, $QP \parallel AC$.
30.03.2018 10:14
just use angle chasing
30.03.2018 14:03
It's just Pascal's theorem for $(BBMACD)$
01.09.2018 17:29
Let $T=MA \cap CD$. Obviously $BD$ is a diameter of the circle. So, $\angle QCT = \angle QCD = \angle QAD = \angle QAD = \angle MAD$ and $\angle BMC = \angle BMA$ $\implies $ $P$ is the incenter of $ \triangle QCM$ $\implies $ $\angle CQP = \angle PCT $ And $$-1=(DB, AC) \overset{M} = (DP, TC)$$So by Right angles and bisectors theorem we get $ \angle PQD = 90^{\circ}$.So $PQ \bot BD$ but $AC \bot BD$ SO, $PQ \parallel AC$
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27.07.2019 19:47
$\angle QDP=90^{\circ}-\angle CBD=\angle BCA=\angle PMQ \implies DQPM $ is cyclic $\implies$ By Reim's Theorem $\implies PQ ||AC$
27.07.2019 19:56
$\angle{AMB}=\angle{ADB}=\angle{CDB}$ so $PMDQ$ is cyclic. Now $\angle{DQP}=180^\circ-\angle{BMD}=90^\circ=\angle{DXC}$ and done
22.11.2019 15:23
Was the constriant $AD=DC$ required at all? anantmudgal09 wrote: Let $ABCD$ be a cyclic quadrilateral with $AB=BC$ and $AD = CD$. A point $M$ lies on the minor arc $CD$ of its circumcircle. The lines $BM$ and $CD$ meet at point $P$, the lines $AM$ and $BD$ meet at point $Q$. Prove that $PQ \parallel AC$. As $AB=BC$, hence, $B$ is the midpoint of $\widehat{AC}$. Hence, $AC\|BB$. Let $P_{\infty}$ be the line along $AC$. So, now the conclusion follows by just applying Pascal on $BDCAMB$. So, $PQ\|BB\|AC$. Hence, proved. $\blacksquare$.
26.02.2020 10:40
Can we say that $ABCD$ is a rhombus? Oh no, just realized we can't.
16.11.2021 15:49
Quite easy Notice that $ABCD$ Is a kite and is cyclic so $\measuredangle DAB = \measuredangle DCB = 90^{\circ}$ and $BD \perp AC$. Let $O=BD \cap AC$ Now $$\measuredangle BAM + \measuredangle DBA = \measuredangle BAO + \measuredangle KAM + \measuredangle OBA = \measuredangle CBF + 90^{\circ} = \measuredangle CBF + \measuredangle FCB \implies \measuredangle AGB = \measuredangle BFC \implies \measuredangle EGD = \measuredangle EFD$$which means $\odot(DGFE)$ and we are done by Reims theorem $\blacksquare$
23.12.2021 10:29
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25.12.2021 02:38
Solution 1: (By angles). $\angle BCA=\angle BAC=\angle CDB=\angle BDA=\angle BMA$ $\implies PMDQ$ is cyclic then: $\angle PQM=\angle PDM=\angle CAM$ $\implies PQ \parallel AC$$\blacksquare$ Solution 2: (By harmonic) Let: $CD \cap AM=L$ $\implies (A,C;B,D) \overset{M}{=}(L,C;P,D)\overset{Q}{=}(A,C;PQ \cap AC,DQ \cap AC)$ but $DQ \cap AC$ is the midpoint of $AC$ $\implies PQ \parallel AC$$\blacksquare$
25.12.2021 02:56
Can we use Pascals here?
09.02.2022 10:20
Note $C$ is the reflection of $A$ in $\overline{BD}$ and $\angle BCD=90.$ Hence, $$\measuredangle PDQ=\measuredangle BDA=\measuredangle BMA$$so $PQDM$ is cyclic. Therefore, $$\measuredangle DQP=90=\measuredangle (\overline{DB},\overline{AC}).$$$\square$
24.06.2022 09:46
19.12.2022 22:12
Note that $\angle QMP = \angle AMB = \angle ADB = \angle BDC = \angle QDP$ so $QPDM$ is cyclic. Thus, $\angle PQD=90^\circ$ so $PQ\perp BD.$ Since $BD\perp AC$, $AC\parallel PQ$ as desired.
23.03.2024 21:51
A little fun one to get the blood flowing. We use directed angles $\measuredangle$ modulo $\pi$. Claim: $PQDM$ is cyclic. Proof: We have that \begin{align*} \measuredangle PMQ &= \measuredangle BMA \\ &= \measuredangle BCA \\ &= \measuredangle CAB \\ &= \measuredangle CDB \\ &= \measuredangle PDQ. \phantom{e} \square \end{align*} From here, we have that \begin{align*} \measuredangle PQM &= \measuredangle PDM \\ &= \measuredangle CDM \\ &= \measuredangle CAM, \end{align*}so that $AC \parallel QP$, as desired. $\blacksquare$
02.07.2024 21:12
Easy angle chase
03.07.2024 00:09
Clearly $DQPM$ is cyclic as $\angle BDP = \angle QMB$. And since $BD$ is a diameter of $ABCD$ we have $\angle BMD = 90^\circ = \angle PQD$ which implies the desired as both $PQ$ and $AC$ are perpendicular to $BD$.