Denote E,F,G,H be the orthogonal projs of R on AB,BC,CD,DA respectively. With mod $ \pi$: (EH,ER)=(PH,PR), (ER,EF)=(BR,BF), so (EH,EF)=(PH,PR)+(BR,BF) Similarly, (GH,GF)=(DH,DR)+(QR,QF) We knew that \angle PRD=\angle QRB Hence (EH,EF)=(GH,GF), then E,F,G,H are concyclic. Note: If quadrilateral ABCD is cyclic, E,F,G,H are collinear.

mr.danh wrote: Note: If quadrilateral ABCD is cyclic, E,F,G,H are collinear. More exactly, when $ ABCD$ is cyclic, $ R$ coincides with the Miquel point of the complete quadrilateral $ ABCDEF$ and therefore $ E$, $ F$, $ G$, $ H$ all lie on the so called Simson line of $ ABCD$, which is parallel to the Aubert line and perpendicular to the Gauss line of the given quadrilateral.

With mod \pi: (EH,ER)=(PH,PR), (ER,EF)=(BR,BF), so (EH,EF)=(PH,PR)+(BR,BF) what does it mean？sorry for my unwisdom

plane geometry wrote: With mod \pi: (EH,ER)=(PH,PR), (ER,EF)=(BR,BF), so (EH,EF)=(PH,PR)+(BR,BF) what does it mean？sorry for my unwisdom Oriented angles,I guess...

any solutions without Oriented angles please

Any solution please

Dear Mathlinkers, a similar problem intitled "The Neuberg-Mineur circle" can be found on my website http://perso.orange.fr/jl.ayme Sincerely Jean-Louis

please any solution

Lemma China TST 2002 Day 1 problem 1, $ ABCD$ is quadrilateral, define $ AB \cap CD = P$, $ AC \cap BD = O$, $ AD \cap BC = Q$, orthogonal projection of $ O$ on the line is $ R$. Then, $ \angle ARC = \angle BRD$ If we define $ X, Y, Z, W$ be orthogonal projections of $ R$ on the line $ AB, DC, DA, AB$ respectively, then it is easy to show $ \angle XYW = \angle XZW$ with lemma:)

Sorry to revive, but the above post contains an error; the second $AB$ in the last line should actually be a $BC$. Also, the $ARC$ and $BRD$ should actually be $BRC$ and $ARD$.

plane geometry wrote: With mod \pi: (EH,ER)=(PH,PR), (ER,EF)=(BR,BF), so (EH,EF)=(PH,PR)+(BR,BF) what does it mean？sorry for my unwisdom Basically, they're angles. The common point at the left is the vertex. For example, $(EH,ER)=(\angle HER)$. I think that there is a typo here: $H$ and $G$ are switched.

@above, so does order matter? (i.e. are $(EH, ER)$ and $(ER, EH)$ the same thing?)

I noticed that $GE$ and $HF$ intersect on $RO$. Did anyone manage to prove something like this?

Heebeen, Yang wrote: Lemma China TST 2002 Day 1 problem 1, $ ABCD$ is quadrilateral, define $ AB \cap CD = P$, $ AC \cap BD = O$, $ AD \cap BC = Q$, orthogonal projection of $ O$ on the line is $ R$. Then, $ \angle ARC = \angle BRD$ If we define $ X, Y, Z, W$ be orthogonal projections of $ R$ on the line $ AB, DC, DA, AB$ respectively, then it is easy to show $ \angle XYW = \angle XZW$ with lemma:) How is it easy?I am not getting it

Pluto1708 wrote: How is it easy?I am not getting it while it's not obvious, it is easy in the sense that only some simple angle chasing will be needed to complete the proof

Let $W,X,Y,$ and $Z$ be the orthogonal projections of $R$ onto $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively. Lemma: (China TST 2002 D1 P1) Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD.$ The two diagonals meet at $P.$ Let $O$ be the foot of the perpendicular from $P$ to $EF.$ Show that $\angle BOC=\angle AOD.$ Proof. Since $$-1=(C,D;\overline{FP}\cap\overline{BC},E)\stackrel{F}{=}(C,A;P,\overline{CP}\cap\overline{FE})$$and $\angle POE=90,$ we know that $\overline{OP}$ bisects $\angle COA.$ Similarly, $\overline{OP}$ bisects $\angle BOD$ and we are done. $\blacksquare$ Hence, \begin{align*}\measuredangle YZX&=\measuredangle WZA+\measuredangle QZX\\&=\measuredangle WRA+\measuredangle QRX\\&=\measuredangle CRX+\measuredangle WRP\\&=\measuredangle CYX+\measuredangle WYC\\&=\measuredangle WYX.\end{align*}$\square$