Classic humpty point problem Let $T_A$ be the $A$-humpty point of $\triangle ABC$. Let $A'$ be the feet of altitude from $A$ to $BC$, let $A$ altitude intersect circumcircle at $H_1$ and $H_2$ be $A$-antipode. Let $EF \cap BC = Z$. Claim. $Z,H,T_A$ collinear (well known). Proof. Suppose $MH$ intersect circumcircle at $K$ and $H_2$. Notice that \[ \measuredangle KAA' \equiv \measuredangle KAH = \measuredangle KAH_1 = \measuredangle KH_2 H_1 = \measuredangle KMA' \]Hence, $KAMA'$ is cyclic. First, take radical axis on $(BFEC), (ABC), (AKHFE)$. This gives us $AK, EF, BC$ concur. So $A,K,Z$ collinear. Now, radical axis on $\Gamma, (AKHFT_A E), (AKA'M)$ gives us $AK,HT_A, A'M$ concur. Therefore, $H,T_A, AK \cap A'M = Z$ concur. Claim. $Z,X,Y$ collinear. Let $HM \cap T_A A' = L$. By Brocard Theorem, $ZL$ is the polar of $A$. Similarly, $XY$ is the polar of $A$. Thus, $Z,X,Y,L$ lies on the polar of $A$, implying the result. We finish off by noticing that \[ ZB \cdot ZC = ZF \cdot ZE = ZH \cdot ZT_A = ZX \cdot ZY \]

Let $\odot(HM)$ intersect $AM$ and $BC$ at points $H_A$ and $D$ respectively. Simply $D$ is the foot of perpendicular from $A$ and let points $E,F$ be the foots of perpendiculars from points $B,C$ respectively. Let $HH_A$ intersects the line $BC$ at point $T$. Claim: Points $T,X,Y$ are collinear. Proof: By Brocard's theorem we have that $T$ lies on a polar of $A$. Observe that $XY$ is clearly also a polar of $A$. Thus points $T, X, Y$ are collinear as desired. Claim: Line $EF$ also passes through point $T$ Proof: Consider inversion centred at point $A$ with radius $\sqrt{AH \cdot AD}$. Under this inversion points $B, C, H, H_A$ maps to points $F, D, M, E$. This implies $\odot(FDME)$ maps to $\odot(BHH_AC)$, therefore quadrilateral $BHH_AC$ is cyclic. Now by radical axis theorem on $\odot(AFHH_AE)$, $\odot(BHH_AC)$ and $\odot(BFEC)$ we have that lines $EF, BC, HH_A$ are concurrent at point $T$. Now PoP finishes the problem: $$ TF \cdot TE = TH_A \cdot TH = TX \cdot TY = TB \cdot TC$$This proves that quadrilateral $BXYC$ is cyclic as desired.

Define $D$, $E$ and $F$ as the feet of the perpendiculars from $A$, $B$ and $C$ respectively. Let $BC$ meet $EF$ at $K$, and let $AK$ meet $(ABC)$ at $P$. By power of a point, $KP.KA=KB.KC=KE.KF$ therefore $AEFP$ is cyclic, and, more importantly, $P$ lies on the circle with diameter $AH$. This means that $HP \perp AP$. Let $A'$ be the antipode of $A$. Once again, $A'P \perp AP$, hence $P$, $H$ and $A'$ are colinear, and a corollary is that $M$ also lies on this line. Let $KH$ meet $AM$ at $L$. We have $AD \perp MK$ and $MP \perp AK$, and they intersect at $H$, therefore $H$ is the orthocentre of $AMK$, hence $KL \perp AM$. A corollary is that the points $L$, $M$, $D$ and $H$ lie on a circle with diameter $HM$, which is $\Gamma$. Furthermore, L lies on the circle $(AEHFP)$. Let $O$ be the centre of $\Gamma$, or the midpoint of $HM$. As $AX$ and $AY$ are tangents to $\Gamma$, $AX \perp XO$ and $AY \perp YO$. Furthermore, as $AP \perp HP$, $AP \perp OP$ hence $AXOYP$ is a cyclic pentagon with diameter $AO$. More importantly, $AXYP$ is cyclic. Consider the circles $(APXY)$, $(ALHP)$ and $(LXYH)$. These circles have radical axes $AP$, $LH$ and $XY$, and therefore they all concur at a point, namely $K$. By power of a point from $K$, we have $KX.KY=KP.KA=KB.KC$, therefore $BXYC$ is cyclic.

Coaxility lemma FTW: $DEF$ is the orthic triangle. Let $O$ be the midpoint of $HM$. $(AXOY)$ and $(XYMD)$ are two circles passing through $X,Y$. Thus it is enough to prove that $\frac{P_{(XYDM)}(B)}{P_{(XYDM)}(C)}=\frac{P_{(AXOY)}(B)}{P_{(AXOY)}(C)}$. Now observe that $$\frac{P_{(XYDM)}(B)}{P_{(XYDM)}(C)}=\frac{BD\cdot BM}{CD \cdot CM}=\frac{BD}{CD}\overset {Ceva}=\frac{BF \cdot AE}{CE \cdot AF}=\frac{BF \cdot AB}{CE \cdot AC}=\frac{\frac{3}{4} BF \cdot \frac{}{}AB}{\frac{3}{4} CE \cdot AC}=\frac{P_{(AXOY)}(B)}{P_{(AXOY)}(C)}$$$\square$

Let $D,E,F$ be the feet of the altitudes for $A,B,C$ and let $N$ be the humpty point of $\triangle ABC$. It is well known that $EF,HN,BC$ concur, call it $P$. It is easy to see that $AFHNE$ and $HDMN$ are cyclic. So $\Gamma = (HDMN)$. Note that by Brocard's Theorem, $P$ lies on the polar of $A$ w.r.t to $\Gamma$. Also, since $AX,AY$ are tangents, $XY$ is the polar of $A$, hence $P,X,Y$ are collinear. Thus, $PX \cdot PY = PH \cdot PN = PF \cdot PE = PB \cdot PC$, we are done.

Let $H_a$ be the $A$-Humpty point of $\triangle ABC$, it's well-known that $H_a$ lies on $AM$ and $H_a$ lies on $(HM)$. Let $D$ be the foot from $A$ to $BC$. Then by Brocard $E=HH_a\cap DM$ lies on the polar of $A$, which is $XY$. Now $$EX\cdot EY=ED\cdot EM=EB\cdot EC,$$where the last equality is McLaurian Identity. Done.

Let $HM$ meet circumcircle of $ABC$ at $Z$.Let $AD$ be altitude and Let $O$ be center of circle with diameter $HM$. Claim1 : $AZDM$, $AZXOY$ and $HXDMY$ are cyclic. Proof : Note that $AZM = \angle 90$ so $\angle AZM = \angle ADM$ so $AZDM$ is cyclic. $\angle AXO = \angle AYO = \angle AZO = \angle 90$ so $AZXOY$ is cyclic. $\angle HDM = \angle 90$ so $D$ lies on circle with diameter $HM$ so $HXDMY$ is cyclic. Now by Radical Axis Theorem we have $AZ$,$XY$ and $DM$ are concurrent at $S$. we have $SX.SY = SZ.SA = SB.SC$ so $BXYC$ is cyclic. we're Done.

Let $AD,BE,CF$ be altitudes, $S$ be $\text{A-Humpty point}$, $J$ be midpoint of $HM$ and $R$ be second intersection point of $(ABC)$ and $(AEF)$. It's well-known that $AR-EF-HS-BC$ are concurrent and let this concurrency point be $T$. Also we know that $R-H-M$ are collinear.$\angle AYJ=\angle ARJ=90 \implies ARJY$ is cyclic, so $ARJX$. So we get $ARJXY$ is cyclic. From Radical axises of $(ARHS),(HSXY),(XYAR)$ we get $XY$ passes through $T$. So $TX\cdot TY=TH\cdot TS=TB\cdot TC \implies BCXY$ is cyclic. So we are done.

Let $H_A$ be the $A$-HM point of $\triangle ABC,$ let $T=\overline{BC}\cap\overline{HH_A},$ and let $D$ be the foot from $A$ to $\overline{BC}.$ Notice $H_A$ lies on $(HBC)$ and $\Gamma.$ By Brocard's Theorem on $HH_AMD,$ $T$ is on the polar of $A$ with respect to $\Gamma$ so $T$ lies on $\overline{XY}.$ Then, Radical Axis on $\Gamma,(BHH_AC),$ and $\{B,C,X,Y\}$ finishes. $\square$

IndoMathXdZ wrote: Given an acute triangle $ABC$, let $M$ be the midpoint of $BC$ and $H$ the orthocentre. Let $\Gamma$ be the circle with diameter $HM$, and let $X,Y$ be distinct points on $\Gamma$ such that $AX,AY$ are tangent to $\Gamma$. Prove that $BXYC$ is cyclic. Let $P$ be the $A-HUMPTY $ point and $T=BC\cap EF$ then $T,H,P$ are colinear. By Brocard's Theorem on $HPMD.AT$ we have that$T,X,Y$ are colinear. Then by power point we have $TX.TY=TH.TP=TB.TC$

Let $D$ be the foot of the $A$-atltidue, $H_A$ be the $A$-Humpty point, and $X_A = \overline{HH_A} \cap \overline{BC}$. Note that $H_A$ is the foot from $H$ onto $\overline{AM}$ so it lies on $\Gamma$, and $D$ lies on $\Gamma$ too. Then by Brocard we find that $X_A$ lies on the polar of $A$ wrt $\Gamma$, or line $XY$, so $X, Y$, and $X_A$ are collinear. To finish, note that $H_A$ lies on $(BHC)$ as $A' = 2M - A$ is the $H$-antipode of $(BHC)$, so $$X_AB \cdot X_AC = X_AH \cdot X_AH_A = X_AX \cdot X_AY,$$which finishes.

Let $D,E,F$ denote the feet of the altitudes from $A,B,C$ respectively. Note that $D$ lies on $(HM)$. Let $Z=XY \cap BC$, and $G=AM \cap (HM)$. Note that $XY$ is the pole of $A$ w.r.t. $(HM)$, so $HG \cap DM$ lies on $XY$ by Brocard's theorem. But $Z=DM \cap XY$, so $Z,H,G$ collinear. We claim that $BHGC$ is concyclic. Taking an inversion about the circle centred at $A$ with radius $\sqrt{AH \cdot AD}$, the points $B,H,G,C$ go to $F,D,M,E$ respectively, which all lie on the nine-point circle. It follows that $BHGC$ is concyclic. By power of point, $ZX \cdot ZY = ZH \cdot ZE = ZB \cdot ZC$, so $BCXY$ concyclic. $\textit{Remark: } G$ is the $A$-Humpty point of $\Delta ABC$.

Let $H_A$ be the $A$-humpty point, $Q$ be the $A$-queue point, and $P=\overline{AQ} \cap \overline{HH_A} \cap \overline{BC}$; note that $\Gamma=(HMH_A)$. Invert about $M$ with radius $MB=MC$. This swaps $Q$ and $H$ as well as $A$ and $H_A$, so $X$ and $Y$ are sent to the points $X',Y'$ on $\overline{APQ}$ such that $(MH_AX')$ and $(MH_AY')$ are tangent to $\overline{APQ}$. Then we want to show $$PA\cdot PQ=PB\cdot PC=PX'\cdot PY'=PA^2-PX'^2=PA^2-PH_A\cdot PM \iff AQ\cdot AP=AH_A\cdot AM,$$which follows from the fact that $PQH_AM$ is cyclic since $\angle PH_AM=\angle PQM=90^\circ$. $\blacksquare$

Help with this question \[\text{A triangle }ABC\text{ has its centroid as }G\text{. Let }D\text{ be a point on the line parallel to the line }\overline{BC}\text{ that passes through }A\text{ such that }\angle CGD=90^o\text{. Prove that }AB\cdot DC\geq2[AGCD]\]

Solved with jeshts. Construct $P$ being the $A-$Humpty point, $D$ being the projection of $A$ to $BC$. Define $R$ and $S$ the intersection of $XY$ with $AD$ and $AM$ respectively. Since $AX$ and $AY$ are tangents to $\Gamma$, notice that $\overline{AD}$ is the $D-$symmedian of $\Delta XDY$. Therefore, $(D, H; R, A) = -1$. Similarly, since $\overline{AM}$ is the $M-$symmedian of $\Delta XMY$, we have $(M, P; S, A) = -1$. Thus, $RS$, $DM$, $HP$ concur at $K$. Afterwards, notice that $KX \cdot KY = KH \cdot KP$ and since $HPCB$ is cyclic, $KH \cdot KP = KB \cdot KC = KX \cdot KY$, and thus $XYCB$ is cyclic.

Quickly did that Let $D,E$ and $F$ be altitudes feet from $A,B$ and $C$ to $BC,CA$ and $AB$. $MH \cap (ABC)$ at $Q$. So, $AQ,BC$ and $EF$ concurr at point $G$. Let $GH \cap AM$ at $H_a$. Look at $EFCB$ its compelte quadrilateral is $AGC$. From brokard's theorem $G$ lies on polar of $A$ similarly look that $XY$ also lies on polar of $A$. Thus, $G-X-Y$. From $HH_aCB$, $AQHH_a$ and $FECB$ cyclic we found that $G$ intersecrion of radical axises which concluded by $HH_aYMDX$ cyclic $\implies$ $BXYC$ cyclic.

Oops got a hint for humpty by accident Introduce the $A$-Queue point of $\triangle ABC$ as $Q_a$ and the $A$-humpty point as $H_a$. The fact that $AY$ and $AX$ are tangent to $(HM)$ implies that $AXYD$ cyclic with $AD$ diameter where $D$ is the midpoint of $HM$. Since $\measuredangle AQ_aD=\measuredangle AQ_aH=90^{\circ}$ $Q_a\in(AXYD)$, so by radical axis $AQ_a$, $HH_a$, $XY$ concur. But then $BC$ also goes though this line which is well known and so by the converse of the radical axis theorem we must have $BXYC$ cyclic.