The external bisectors of the angles of the convex quadrilateral $ABCD$ intersect each other in $E,F,G$ and $H$ such that $A\in EH, \ B\in EF, \ C\in FG, \ D\in GH$. We know that the perpendiculars from $E$ to $AB$, from $F$ to $BC$ and from $G$ to $CD$ are concurrent. Prove that $ABCD$ is cyclic.
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Tags: geometry, Romanian TST, TST, cyclic quadrilateral
16.05.2021 22:25
Let $P=\overline{AB}\cap\overline{CD}$, and let $Q=\overline{AD}\cap\overline{BC}$. WLOG, assume that $Q,B,C$ and $P,D,C$ lie in these orders. Then since $E$ and $G$ are the incenter and $Q$-excenter of $\triangle QAB$ and $\triangle QBD$ respectively, it follows that $Q,E,G$ are collinear. Similarly, $P,F,H$ are collinear. Let these two lines intersect at $I$. We will now show that $I$ and $T$ are isogonal conjugates with respect to quadrilateral $EFGH$. Notice that $\overline{ET}$ and $\overline{EI}$ are isogonal in $\angle E$, as $\overline{EI}$ passes through the circumcenter of $\triangle EAB$ by the incenter lemma. Similarly, $\{\overline{FT},\overline{FI}\}$, and $\{\overline{GT},\overline{GI}\}$ are also isogonal in $\angle F$ and $\angle G$. If $d(X,l)$ denotes the distance of point $X$ from line $l$, we have \[\frac{d(I,EH)}{d(I,EF)}=\frac{d(T,EF)}{d(T,EH)}\]and similarly for $F$ and $G$. Multiplying them gives us \[\frac{d(I,EH)}{d(I,GH)}=\frac{d(T,GH)}{d(T,EH)}\]which shows that $\overline{HI}$ and $\overline{HT}$ are isogonal in $\angle H$ as well. Therefore, $I$ and $T$ are isogonal conjugates. But this implies that \[\angle EIF+\angle GIH=180^\circ \Longrightarrow EG\perp FH.\]The rest can be done with angle chasing, but there is an easier way to see it. Notice that $\overline{BC}$ is obtained by reflecting $\overline{AD}$ over $\overline{QI}$. Since $\overline{PI}$ is perpendicular to $\overline{QI}$, reflecting $\overline{AD}$ over $\overline{PI}$ gives a line with the same direction as $\overline{BC}$. i.e. $AD$ and $BC$ are anti-parallel in $\angle P$, and the conclusion follows. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(150); pair B = dir(200); pair C = dir(340); pair D = dir(70); pair Q = extension(A,D,B,C); pair P = extension(A,B,C,D); pair E = incenter(Q,A,B); pair H = incenter(P,A,D); pair G = 2*circumcenter(incenter(Q,D,C),D,C)-incenter(Q,D,C); pair F = 2*circumcenter(incenter(P,B,C),B,C)-incenter(P,B,C); pair I = extension(E,G,F,H); pair T = extension(E,foot(E,A,B),F,foot(F,B,C)); draw(A--B--C--D--cycle, black+1); draw(A--Q); draw(B--Q); draw(A--P); draw(D--P); draw(Q--G); draw(P--F); draw(circumcircle(A,B,C), mydash); draw(E--F--G--H--cycle, royalblue+1); draw(E--T, mydash); draw(F--T, mydash); draw(G--T, mydash); draw(H--T, mydash); dot("$A$", A, dir(A)); dot("$B$", B, dir(225)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(135)); dot("$F$", F, dir(270)); dot("$G$", G, dir(0)); dot("$H$", H, dir(30)); dot("$I$", I, dir(135)); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(180)); dot("$T$", T, dir(60)); [/asy][/asy]
16.05.2021 23:36
Claim: Let the perpendicular from $F$ to $BC$ and $E$ ta $AB$ meet at $I$.Then unconditionally $EFGH$ is cyclic quadrilateral and $IE=IF$. Proof. $\angle FBC=90^{\circ}-\frac{\angle B}{2}$.Similarly $\angle EBK=90^{\circ}-\frac{\angle B}{2}$.Since $IF\perp BC$ and $IE\perp AB $ hence $\angle IEK=\angle IFE\implies IE=IF$. Concyclicity of $EFGH$ can be seen by observing that $\angle BFC= \frac{\angle B}{2}+\frac{\angle C}{2}$ and $\angle AHD= \frac{\angle A}{2}+\frac{\angle D}{2}$. So $\angle E+\angle H=1/2(\angle B+\angle C+\angle A+\angle D)=180^{\circ}$ $\blacksquare$ Now it is given that perpendicular from $G$ to $CD$ passes through $I$ which means $IE=IF=IG$.Hence,$I$ is the center of $\odot(EFGH)$.So $IH=IE$. 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Proof. .Let the 2 internal bisector meet each other at $L$. We may assume that $BA$ and $CD$ are not parallel.(If they are parallel the claim can be check easily)Also we may assume $BA$ and $CD$ meet at $X$ so that $X$ and $BC$ lies in the opposite side of $AD$.The other case is analogus.2\angle So observe that $H$ is the incenter of $XAD$ ,$L$ is the incenter of $XBC$ and $F$ is the $X$-excenter of $XBC$.So $H,L,F$ lies on the angle bisector of $\angle BXC$. $\blacksquare$ Now observe that $LBFC$ is concyclic.Indeed $\angle LBF=\angle LCF=90$.This gives us $\angle LFB=\frac{\angle C}{2}$. Now note that,$\angle HIE=2\angle HFE=2\angle LFE=\angle C$. So, $$180^{\circ}=\angle IEH+\angle IHE+\angle HIE=2\angle IEH+\angle HIE=2[90^{\circ}-\angle EAK]+\angle C=\angle A+\angle C$$Hence $ABCD$ is cyclic quadrilateral. $\blacksquare$
18.05.2021 13:15
Let $AB\cap CD=Q,\ AD\cap BC=P$ and internal angle bisectors of $A,B$ meet at $I_1$. Now, as $EHF$ and $QAB$ are orthologic, we have that the perpendicular from $Q$ to $AB$ passes through $I_1$ but observe that this is the angle bisector of $\angle AQB$. Thus, $EI_1\perp HF$. But, in $\Delta EAB$, $EI_1$ passes through circumcenter of $AB$. Thus, if we consider reflection of $EABI_1$ across angle bisector of $AEB$ and scale, we get that we can get $HF$. Thus, $ABHF$ is cyclic. Similarly, $CDHF$ are cyclic. But observe that $AB,CD,HF$ concur at $P$ as $H,F$ lie on the angle bisector of $APD$. But then by power of point from $P$, we get that $ABCD$ is cyclic and we are done.