Let ABCD be a cyclic quadrilateral in which AB = BC and AD =CD. Point M is on the small arc CD of the circle circumscribed to the quadrilateral. The lines BM and CD intersect at point P, and the lines AM and BD intersect at point Q. Prove that PQ is parralel to AC.
Problem
Source: Saudi Arabia JBMO training test 2 2019, P4
Tags: geometry
19.04.2020 23:41
Since $ABD$ is cyclic and $AB=CB$ and $AD=CD$, we know that $ABCD$ must be a kite and $BD$ is the diameter of $(ABCD)$. [asy][asy] import olympiad; import geometry; size(200); currentpen=fontsize(9); pair a,b,c,d,p,q,m; b=dir(90); d=dir(-90); a=dir(90+64); c=dir(90-64); draw(a--b--c--d--a); draw(unitcircle); m=dir(-50); p=extension(b,m,c,d); q=extension(a,m,b,d); draw(b--d--m^^a--c^^a--m--b^^p--q); label("$A$",a,dir(a)); label("$B$",b,dir(b)); label("$C$",c,dir(c)); label("$D$",d,dir(d)); label("$P$",p,dir(-5)); label("$Q$",q,dir(210)); label("$M$",m,dir(m)); pair x = extension(a,c,b,d); label("$X$",x,dir(90+45)); draw(circumcircle(q,m,p),dashed+linewidth(0.5)); [/asy][/asy] Let $X=AC\cap BD$. We wish to prove that $PQ\parallel AC$, which is equivalent to $\angle DQP=\angle DXC$. However, note that $\angle DXC=90^\circ$ because the diagonals of a kite are perpendicular to each other. First, we will show that $QPMD$ is cyclic. Note that, since $AB=BC$, we know that $\widehat{AB}=\widehat{BC}\implies\angle AMB=\angle BDC\implies \angle QMP=\angle QDP$ as desired. Therefore, $\angle DQP+\angle DMP=180^\circ$. But, since $BD$ is the diameter, $\angle DMP=\angle DMB=90^\circ$; this means $\angle DQP=90^\circ$. This shows that $\angle DQP=\angle DXC$, and we are done.
28.02.2023 11:27
Moving Points, why not? Change $M$ to an arbitrary point on $(ABCD)$. Let there be 3 points $M_1,M_2,M_3$ where the statement holds. $P$ and $Q$ would move along $M$, so we check whether they preserve a cross ratio. $B(M_1,M_2;M_3,M)=(P_1,P_2;P_3,P)$ $A(M_1,M_2;M_3,M)=(Q_1,Q_2;Q_3,Q)$, so they both preserve a cross ratio. Now take $P=C, P=D, P=B, P=A$, they all holds. As we checked four cases we're done.
28.02.2023 12:02
We know, $\angle ADB=\angle ACB=\angle CAB$ (since $AB=BC$), $=\angle CDB=\angle AMB$. Hence, quadrilateral $QPMD$ is cyclic. So, $\angle CDM=\angle PDM=\angle PQM=$ Also, $\angle CDM=\angle CAM$. Hence, $PQ$ and $AC$ have equal corresponding angles (with transversal $AM$), which means that they are parallel. $\blacksquare$