First of all let points $M,N$ be symmetric points of $C$ wrt $D,B$. Now easily get $CHTM,CHSN$ cyclic. And let $Y,X$ be the centers of these circles respectively. Now if you try to prove by Menelaus theorem that the perp bisectors of $HS,HT$ intersect on $AH$(which is equivalent to the given problem) you will reduce the problem to $XS/XA=YT/YA$. Now this reduces to $XH/HY=XA/AY$. Now this can be proven if you prove that $AHC$ is the apollonious circle for segment $XY$. If you intersect the perp bisector of $AH$ with $XY$ at $T'$ this reduces to $T'A^2=T'X\cdot T'Y$ now you can angle chase that $T'AX\sim T'YA$ which is trivial.

Here was a sketch of my solution. Not very elegant but quite direct. [asy][asy] size(9cm); pair A = dir(110); pair B = dir(200); pair D = dir(340); pair C = -A; pair H = foot(A, B, D); draw(A--B--D--A--C--H--A); pair M1 = bisectorpoint(C,H); pair O = extension(A, D, M1, midpoint(C--H)); pair T = IP(A--D, CP(O, C)); draw(unitcircle); pair P = extension(midpoint(T--H), O, A, H); draw(CP(O, C)); draw(P--O--H--T); pair M2 = bisectorpoint(C,H); pair O1 = extension(A, B, M2, midpoint(C--H)); pair S = IP(A--B, CP(O1, C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$C$", C, dir(C)); dot("$H$", H, dir(H)); dot("$O$", O, dir(O)); dot("$T$", T, dir(T)); dot("$P$", P, dir(P)); dot("$S$", S, dir(S)); /* Source generated by TSQ */ [/asy][/asy] First by angle chasing one can show that $\angle ATH = \angle TCH + 90^{\circ}$, so the tangent to $(CHT)$ at $T$ is perpendicular to $AD$. Thus the circumcenter $O$ of $\triangle TCH$ lies on $AD$. Let the perpendicular bisector of $TH$ meet $AH$ at $P$ now. It suffices to show that $\frac{AP}{PH}$ is symmetric in $b = AD$ and $d=AB$, because then $P$ will be the circumcenter of $\triangle TSH$. To do this, set $AH = \frac{bd}{2R}$ and $AC=2R$. Use the Law of Cosines on $\triangle ACO$ and $\triangle AHO$, using variables $x=AO$ and $r=HO$. We get that \[ r^2 = x^2 + AH^2 - 2x \cdot AH \cdot \frac{d}{2R} = x^2 + (2R)^2 - 2bx. \] By the Angle Bisector Theorem, $\frac{AP}{PH} = \frac{AO}{HO}$, so the rest is a ~15 minute computation now.

we need to proof .

It's some easier problem for $ [IMO3] $.My solution similar to leader's solution, latter write full solution.

Take inversion wrt center C, then inverse the whole figure wrt center H, and we're done.

My solution: Reflect $C$ about $B$ and $D$ to get $E,F$, then $ESHC$ and $FTHC$ are cyclic (this gets rid of the ugly angle condition and makes the diagram better since now $A$ is the circumcenter of $ECF$). Furthermore, $ES=SC$ and $FT=TC$. Now let $M$ be the midpoint of $EF$ and note that $MAH$ is a straight line and perpendicular to $EF$. Thus $EH=HF$. Simple angle chasing gives $\angle SHT=\angle MHF=\angle MHE$, and $HS$ is the external angle bisector of $\angle EHC$ (same for the other side), useful information for $\triangle SHT$. But now what? There is no way to find the other 2 angles of $\triangle SHT$ (indeed, SHT). One very useful way to do this is to construct another triangle where we can find the angles, and show they are similar using length ratios. So you experiment around for a while, and find that it is similar to the triangle formed by sticking the isosceles triangles $EHF,ESC,FTC$ together. Now you are almost done, just need to formalise a bit. Construct $C_p$ such that $\triangle ESC\sim\triangle EHC_p$ (by spiral similarities $\triangle ESH\sim\triangle ECC_p$). Now $HC_p=HE=HF$ and $\angle EHF+\angle ESC+\angle FTC=360$ (since $\angle ESC=\angle EHC$ and $\angle FTC=\angle FHC$). So we also have $\triangle FHC_p\sim\triangle FTC$. Note that $\angle EC_pF=\frac12\angle EHF=\angle SHT$, and the length ratio $\frac{HS}{HT}=\frac{HS}{HE}\frac{HF}{HT}=\frac{CC_p}{EC_p}\frac{FC_p}{CC_p}=\frac{FC_p}{EC_p}$. Thus $\triangle SHT\sim\triangle FC_pE$, and the rest is just angle chase.

Base on very good idea of leader, I have general problem and solution Problem. Convex quadrilateral $ABCD$ is cyclic. Point $P$ is a point inside triangle $ADB$ such that $\angle PAD=\angle CAB$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $P$ lies inside triangle $SCT$ and $\angle CPS - \angle CSB = \angle TPC - \angle DTC = 90^{\circ}$. Prove that circumcenter of triangle $PST$ lie on $AP$. Solution (It is from idea of leader). Let $M,N$ be reflection of $C$ through $AB,AD$. We have $CPSM$ and $CPTN$ are cyclic with center $X,Y$. We must prove that perpendicular bisector of $PS,PT$ and $PA$ are concurrent. Apply Menelaus theorem for triangle $PST,PTA$ we must prove that $\frac{XS}{XA}=\frac{YT}{YA}$ or $\frac{AX}{AY}=\frac{PX}{PY}=\frac{CX}{CY}.$ Let $XY$ cuts perpendicular bisector of $PA$ at $Q$ and $XY$ cuts $AC$ at $R$. We see \[\angle XAQ=\angle PAQ-\angle PAX=(90^\circ-\frac{\angle AQP}{2})-\angle CAD=90^\circ-\angle ACP-\angle CAD=\angle YRC-\angle CAD=\angle AYX.\] From this we have $QA$ is tangent of $(AXY)$ but $Q$ is circumcenter of triangle $APC$ so $(APC)$ is Apollonious circle for segment $XY$. We are done.

Here is similar to v_Enhance's solution. The same in the beginning but more beautiful bush in the end if i can say so Let $O_1$ and $O_2$ be the centers of the circles. We need to prove that \[\frac{AO_2}{O_2H}=\frac{AO_1}{O_1H}.\] Let $N$ be he midpoint of $HC$. Then $\angle{CNO_2}=\angle{CDA}=\angle{CNO_1}=\angle{CBA}=90^\circ$. Thenquadrilaterals $O_1BNC$ and $O_2DCN$ are inscribed. Since that $\frac{O_2C}{CO_1}=\frac{\sin{O_2O_1C}}{\sin{O_1O_2C}}=\frac{\sin{NBC}}{\sin{NDC}}$. Also $\frac{AO_2}{AO_1}=\frac{\sin{AO_1O_2}}{\sin{AO_2O_1}}=\frac{\sin{BCN}}{\sin{DCN}}$. As $O_2C=O_2H, O_1C=HO_1$ it is sufficient to prove that $\frac{\sin{NBC}}{\sin{NDC}}=\frac{\sin{BCN}}{\sin{DCN}}$ or $\frac{\sin{NBC}}{\sin{BCN}}=\frac{\sin{NDC}}{DCN}$. Finally $\frac{\sin{NBC}}{\sin{BCN}}=\frac{NC}{BN}=\frac{NC}{DN}=\frac{\sin{NDC}}{DCN}$ ($BN=DN$ because $NE \parallel AH$ ($HN=NC, AE=EC$) so $NE \perp BD$ where $E$ is the center of the circumcircle of $ABC$. What do you think about problem? As for me it was really surprisingly to see geometry with angle equalities and without many circles on the 3rd position. And i hope leaders wrote "$H$ lies inside triangle $CST$" in bold - i waste a lot of time finding what is wrong

let $K,L$ on $BC$ such that $BC=BK$ and $CD=DL$. let $HS,BC$ meet at $I$. $HT,CL$ meet at $J$. $CHSK$ and $CHTL$ are cycle. $A$ is circumcenter of $CKL$. so $HK=HL$. $S$ is midpoint of arc $KC$ from circle $CHSK$ so $\frac{IK}{IC}=\frac{HK}{HC}=\frac{HL}{CH}=\frac{JL}{JC}\Rightarrow IJ\parallel KL\parallel BD$ so $\widehat{DHT}=\widehat{HJI}$ $HS.HI=HC^{2}=HT.HJ\Rightarrow \widehat{HJI}=\widehat{HST}=\widehat{DHT}$ so we are done

First note that $(CHS)$ is orthogonal to $AB$, so after inverting with center $H$ the problem becomes: Let $ABD$ be a triangle and $H$ be the foot of the perpendicular to $BD$ from $A$. $C$ is a point on $ABD$ such that $(AHC)$ is orthogonal to $(ABD)$ and $S,T$ are intersections of $CM_D, CM_B$ with the circles with diameter $AB, AD$ respectively, where $M_D, M_B$ are the midpoints of $AB, AD$. Then $ST$ is parallel to $BD$. Indeed, this is true for any $C$ on the circle $\omega$ through $A,H$ orthogonal to $(ABD)$; the center $O_A$ of $\omega$ lies on $M_B M_D$ and $AO_A^2= O_A M_B \cdot O_A M_D$, so that $\omega$ is an $M_D, M_B$ -Apollonius circle of ratio $AB/AD$, which is then the same as $CM_D/CM_B= CS/CT$, as desired.

mathuz wrote: we need to proof . Is this a request for more detail? Notice that \[ r^2 - x^2 = h^2 - 2xh \cdot \frac{d}{2R} = (2R)^2 - 2bx \] where $h = AH = \frac{bd}{2R}$, whence \[ x = \frac{(2R)^2-h^2}{2b - 2h \cdot \frac{d}{2R}}. \] Moreover, \[ \frac{1}{2} \left( \frac{r^2}{x^2}-1 \right) = \frac{1}{x} \left( \frac 2x R^2 - b \right). \] Now, if we plug in the $x$ in the right-hand side of the above, we obtain \[ \frac{2b-2h \cdot \frac{d}{2R}}{4R^2-h^2} \left( \frac{2b-2h \cdot \frac{d}{2R}}{4R^2-h^2} \cdot 2R^2 - b\right) = \frac{2h}{(4R^2-h^2)^2} \left( b- h \cdot \frac{d}{2R} \right) \left( -2hdR + bh^2 \right). \] Pulling out a factor of $-2Rh$ from the rightmost term, we get something that is symmetric in $b$ and $d$, as required.

Again a beautiful geometry problem at IMO problem 3. Initially I thought it is only medium difficulty, but now I think that it is medium to hard. How does it compare to the difficulty of the IMO 2013 geometry problem 3?

First reflect $C$ over $AB,AD$ to get $E,F$, hence we can see that $E\in CB,F\in CD$ and $CHSE,CHTF$ are cyclic quadrilaterals. Next we will prove that $C,H$ are isogonal conjugates wrt $\triangle ATS$. We already have isogonals $AH,AC$, it suffices now to prove $180-\angle THS=\angle H+\angle SCT$(a well known angle condition for isogonal conjugates that lie outside the triangle)$\iff (\angle THF+\angle FHD)+(\angle SHE+\angle EHB)=\angle A+\angle TCH+\angle HCS$. Since $\angle THF=\angle TCF=\angle TFC,\angle SHE=\angle SEC$ we have $\iff \angle A=(\angle SEC-\angle SEH)+\angle EHB+(\angle TFC-\angle TFH)+\angle FHD$ $=(\angle HEB+\angle EHB)+(\angle HFC+\angle FHD)=\angle HBC+\angle HDC$ which is true. Note that $180-\angle THC+\angle STH=\angle TCD+\angle DTC=90$, hence $CH\perp ST$. Denote the circumcenters of $CHS,CHT$ by $J,K$. Let $(J)\cap AB=X,(K)\cap AD=Y$, so $SX,TY$ are the diameters for those circles. Now $BD$ is tangent to $(THS)\iff $the circumcenter of $THS$ lies on $AH\iff $the line through $S,T$ perpendicular to $SH,TH$ respectively meet at $AH\iff \frac {AS}{SX}=\frac {AT}{AY}\iff TS\parallel XY$. However since we've proven $JK\parallel TS$(since $KJ\perp CH,CH\perp TS$) and $K,J$ are the midpoints of $SX,TY$, it follows that $TS\parallel XY$ and we are done.

Dear Mathlinkers, for beginning, did some one explain the construction of the points S and T ? Sincerely Jean-Louis

We are given $\angle{CHS} - \angle{CSB} = 90^{\circ} \implies \angle{CHS} = \angle{CSB}+90^{\circ}$. This is just equivalent to the fact that the tangent to $(CHS)$ at $S$ is perpendicular to $AB$. Similarly, $AD$ is orthogonal to $(CHT)$. Let $O_1,O_2$ be the centres of $(CHS)$ and $(CHT)$ respectively. It is sufficient to show that the centre of $(SHT)$ lies on $AH$. Equivalently, we want to prove that the perpendicular bisectors of $SH$ and $HT$, which are also the angle bisectors of $\angle{AO_1H}$ and $\angle{AO_2H}$ concur on $AH$. By Angle Bisector Theorem, it is sufficient to show that $\frac{AO_1}{O_1H} = \frac{AO_2}{O_2H}$. Let the centre of the circle $(AHC)$ be $X$. From hereon, we angle chase to prove that $(AHC)$ is actually an Apollonius Circle on $O_1O_2$. \[ \angle{O_2AX} = \angle{HAX}-\angle{HAD} = \left( 90^{\circ} - \frac{1}{2} \angle{AXH} \right) - (90^{\circ} - \angle{ADB}) \] \[= \angle{ADB} - \angle{ACH} = \angle{ACB} - \angle{ACH} = \angle{HCB} = \angle{HCS} + \angle{SCB} \] \[= 90^{\circ} - \angle{O_1SC} + \frac{1}{2}\angle{HO_1S} = \frac{1}{2}\angle{SO_1C} + \frac{1}{2}\angle{HO_1S} = \angle{SO_1O_2} \]Therefore, $XA$ is tangent to $\triangle{AO_1O_2}$. Consequently, $XA^2 = XO_1.XO_2$ and hence, $(AHC)$ is the Apollonius Circle. Therefore, $\frac{O_1H}{HO_2} = \frac{O_1A}{AO_2}$.

jayme wrote: Dear Mathlinkers, for beginning, did some one explain the construction of the points S and T ? Sincerely Jean-Louis Lemma $\omega$ is the circumcircle of $\triangle ABC$ with $AB>AC$,and $M$ is the midpoint of $\widehat{BAC}$ of $\omega$.Let $N$ be the pedal of $M$ on segment $AB$,then $BN=CA+AN$. Proof Let $D$ be the pedal of $M$ on $CA$.Since $AM$ is the external angle bisector of $\angle BAC$,we obtain $MN=MD$,and hence $AN=AD$ due to the Pythagorean Theorem.Combining that $MB=MC,\angle MNB=\angle MDC=90^\circ$,it follows that $\triangle MNB\cong \triangle MDC$,then $BN=CD=CA+AD=CA+AN. \Box$ Construction $\angle CHS-\angle CSB=90^\circ\implies \angle CHS=90^\circ+\angle CSB=90^\circ+(90^\circ-\angle BCS)=$ $180^\circ-\angle BCS$.And hence let $C_1$ be the point of symmetry of $C$ with respect to $B$,then $\angle BC_1S=\angle BCS$ due to $AB\perp BC$,so $C,C_1,S,H$ are concyclic.Since $C,C_1,H$ are all fixed,we obtain the location of $S$.Similarly we get $C_2$ and $T$. Main Proof First we prove that $C_1H=C_2H$. Since $AB$ is the perpendicular bisector of $CC_1$ and $AD$ is that of $CC_2$,we obtain that $A$ is the circumcenter of $\triangle CC_1C_2$,and $AH\perp C_1C_2$ due to $AH\perp BD$ and $BD\parallel C_1C_2$,then $C_1H=C_2H$ $\Box$. Secondly we claim that $CH\perp ST$.Denote by $S_1$ the pedal of $S$ on $C_1H$ and similarly define $T_1$.Combining the lemma,we get that \begin{align*}CH\perp ST& \iff CS^2-HS^2=CT^2-HT^2\\& \iff C_1S^2-HS^2=C_2T^2-HT^2\\& \iff C_1S_1^2-HS_1^2=C_2T_1^2-HT_1^2\\& \iff (C_1S_1+HS_1)\cdot (C_1S_1-HS_1)=(C_2T_1+HT_1)\cdot (C_2T_1-HT_1)\\& \iff C_1H\cdot (C_1S_1-HS_1)=C_2H\cdot (C_2T_1-HT_1)\\& \iff C_1S_1-HS_1=C_2T_1-HT_1\\& \iff CH=CH \Box\end{align*} Finally,without loss of generality,we may assume that $\angle BCS=\angle S_1HS=\angle RHS=\alpha,\angle DCT=\angle T_1HT=\angle RHT=\beta$ and $\alpha<\beta.$ Then \begin{align*}\angle AHT& =\angle SHT-(\angle AHR+\angle RHS)\\& =(\alpha+\beta)-[90^\circ-2\alpha-\frac{1}{2}(180^\circ-2\alpha-2\beta)+\alpha]\\& =\alpha\end{align*} which implies that the circumcenter of $\triangle HST$ lies on $AH$,and the conclusion holds. $\Box$

There is another solution that doesnt use ratios, appolonian circles, or construction of new points and has very little angle chasing being very short too. If this post gets has a large number of thanks, ill add it, cause maybe the above solutions are enough idk. Here is a trig solution along the lines of the above posts. Let $C_1, C_2$ be the symmetric points of $C$ wrt $AB, AC$ and note the cyclic quads like above. Since $A$ is the circumcentre of $CMN$, it follows $AH$ is the perpendicular bisector of $C_1C_2$ so $HC_1=HC_2$. Now, take point $S'$ such that $SB \cdot SS' = SC^2$ and note $\angle SHS' = 90$, and obviously $S' \in CHC_1S$, similarly define $T'$. It follows the centre of those circles $S'', T''$ lie on $AB, AC$. The perpendicular bisector of $HS$ is parallel to $HS'$ and if they concur on $AH$ then $\dfrac{AS'}{AT'} = \dfrac{AS''}{AT''} = \dfrac{SS''}{TT''} = \dfrac{r_s}{r_t}$. where $r_s, r_t$ are the radii of the respective circles. Note $\angle SS''T'' = \angle C_1CH$ and similar for $\angle TT''S''$. Now, you can use whatever you want, easiest is sine law cause $\dfrac{\sin HCC_2}{r_t} = \dfrac{\sin HCC_1}{r_s} \implies \dfrac{r_s}{r_t} = \dfrac{\sin AS''T''}{\sin AT''S''} = \dfrac{AS'}{AT'}$ as desired. EDIT: DVDthe first posted it You can avoid ratios though, as $\angle AHT = \angle AS'C = \angle (SH,HC)$ so $HC, AH$ are isogonal.

Consider the transformation $\Psi$ as follows: invert with centre $A$ with radius $\sqrt{AB\cdot AD}$, and reflect about the angle bisector of $\angle BAD$. Note that $\Psi$ maps $B$ to $D$ and $C$ to $H$ (and back too, since $\Psi^2$ is the identity mapping). Reflect $C$ about $B$ and $D$ to get $B',D'$. Now note that $CHB'S(=\omega_1)$ and $CHD'T(=\omega_2)$ are concyclic. Then let $\omega_1\cap AB=S'\neq S$, $\omega_2\cap AD=T'\neq T$. However, $\omega_1,\omega_2$ have centers lying on $AB,AD$ respectively, and both pass through $C$ and $H$. Thus they map to each other under $\Psi$. Hence $AS\cdot AT'=AT\cdot AS'$. Thus $ST$ is parallel to $S'T'$, which then is parallel to $O_1O_2$ (centers of $\omega_1,\omega_2$) Thus $\frac{AO_1}{O_1C}=\frac{AO_2}{O_2C}$, and hence the angle bisectors of $\angle AO_1C$ and $\angle AO_2C$ meet on $AH$, but those are also the perpendicular bisectors of $HS$ and $HT$, so done.

Let $Q$ be the point on $(BCD)$ such that $CQ \parallel BD$, the projection of $C$ onto $BD$ be $R$, $P = CC \cap BD$, and the midpoints of $CB, CD, CP$ be $M, N, K$ respectively. It's clear that $ABCD$ is cyclic with diameter $AC$, $K \in MN$, and $Q \in AH$. Now, we perform $\sqrt{bd}$-inversion wrt $BCD$. It's easy to see $B \leftrightarrow D$, $A \leftrightarrow R$, and $P \leftrightarrow Q$. Moreover, since $CQ \perp \overline{AHQ}$, we know $CPH^*R$ is cyclic with diameter $CP$. Thus, $K$ is the center of $(CPH^*R)$, so equal tangents implies $KH^*$ is tangent to $(BCD)$. Observe that $$\angle CS^*D = \angle CBS = 90^{\circ}$$which means $S^*$ lies on $(CD)$. Hence, $$\angle CS^*H^* = \angle CHS = 90^{\circ} + \angle CSB = 90^{\circ} + \angle CDS^*$$$$= 90^{\circ} + (90^{\circ} - \angle DCS^*) = 180^{\circ} - \angle CS^*N$$so $S^* \in H^*N$. An analogous process gives $T^* \in (CB)$ and $T^* \in H^*M$. Now, since $S$ lies on side $AB$, it's easy to see $S$ lies in the interior of $\angle ACB$. Thus, $S^*$ lies in the interior of $\angle RCD$, which means $S^*$ belongs to the interior of $H^*N$, as $H^*$ is below $BD$. Similarly, we deduce that $T^*$ belongs to the interior of $H^*M$. Because $K = CC \cap MN$, a well-known result yields $(CPH^*R)$ as the $C$-Apollonian circle wrt $MN$. This gives $$\frac{H^*M}{H^*N} = \frac{CM}{CN} = \frac{T^*M}{S^*N}$$so $MN \parallel T^*S^*$, implying $(H^*T^*S^*)$ is tangent to $(H^*MN)$. On the other hand, since $(CMN)$ and $(BCD)$ are tangent, $$KH^{*^2} = KC^2 = | Pow_{(BCD)}(K) | = | Pow_{(CMN)}(K) | = KM \cdot MN$$so $(H^*MN)$ is tangent to $KH^*$. It follows that $(BCD)$, $(H^*MN)$, and $(H^*T^*S^*)$ are pairwise tangent at $H^*$, as required. $\blacksquare$ Remarks: We must view this problem as a configuration in $\triangle BCD$. By Thales', we know $H^* \in AP$. In addition, $$\measuredangle CH^*R = \measuredangle CPR = \measuredangle PCQ = \measuredangle CH^*Q$$or ISL 2011/G4 yields $H^* \in QR$.

I can't believe this works. Apparently, I just walked straight into a solution, but looking back, all the steps are natural enough that I'd call this a reasonably straightforward inversion(s) exercise. Let $T_1$ be the point on $AD$ such that $THCT_1$ is cyclic, and construct $S_1$ similarly. Note that $\angle THC - 90^\circ = \angle TT_1C = \angle T_1HC$, so $\angle THT_1 = \angle TCT_1 = 90^\circ$. Similarly, $\angle SHS_1 = \angle SCS_1 = 90^\circ$. Now, invert at $H$ with power $-HB\cdot HD$. Let $X^*$ denote the image of $X$ under this inversion. Let $P$ be the second intersection of the line $AH$ with the circumcircle of $(ABCD)$. We have that $C^*$ lies on both the circumcircle of $ABCD$ and the line $HC$, so it is the second intersection of $HC$ with $(ABCD)$. This inversion swaps the pairs $(A, P)$, $(B, D)$, $(AD, (HPB))$, $(AB, (HPD))$, $((TT_1CH), T^*T_1^*C^*)$, $((SS_1CH), S^*S_1^*C^*)$. Also, we have that $\angle T^*HT_1^* = 90^\circ$. Together with the fact that $BT^*T_1^*HP$ is cyclic and that $C^*, T^*, T_1^*$ are collinear, this implies that $T^*$ and $T_1^*$ are the two intersections of $(BHP)$ with the line passing through $C^*$ and the midpoint of $PB$. Similarly, $S^*$ and $S_1^*$ are the two intersections of $(DHP)$ with the line passing through $C^*$ and the midpoint of $PD$. Now, consider the inversion at $C^*$ with power $CB\cdot CD$ followed by a reflection about the angle bisector of $\angle BC^*D$. This swaps $(B, D)$, $(H, P)$ (using the well-known fact that $\widehat{BP} = \widehat{CD}$ and that the image of $P$ needs to lie on the $BD$). In particular, this inversion also swaps $(BHP)$ and $(DHP)$. Since $T_1^*$ is the point on $(BHP)$ that is closest to $C^*$ and $S^*$ is the point on $(DHP)$ that is the furthest, $T_1^*$ and $S^*$ swap under this inversion. By similar reasoning, $S_1^*$ and $T^*$ swap. This implies that $C^*T_1^*\cdot C^*S^* = C^*S_1^*\cdot C^*T^*$, so $S_1^*T_1^*\parallel S^*T^*$. Letting $M$ and $N$ as the midpoints of $BP$ and $DP$, we have that $BD\parallel MN\parallel S_1^*T_1^*\parallel S^*T^*$. Inverting back (centered at $H$), we have that $BD$ and $(HST)$ are tangent at $H$, as desired.

Justification of the unicity : We will first prove that there is at most one point $S$ on side $AB$, as well as at most one point $T$ on side $AD$, satisfying the desired angle condition. By way of contradiction, let's pick two such points $S_1$ and $S_2$ on side $AB$, WLOG with $S_1$ closer to $A$ than $S_2$. It's easy to notice that $\angle CHS_1 > \angle CHS_2$, and $\angle CS_1B < \angle CS_2B$. Thus, $\angle CHS_1 - \angle CS_1B > \angle CHS_2 - \angle CS_2B = 90^{\circ}$, and that contradicts the condition. We can use the same argument for $T$. Therefore, that allows us to work backwards. If we manage to construct two points $S'$ and $T'$ such that $(S'HT')$ is tangent to $BD$ and the angle condition is verified, we'll immediately have $S = S'$ and $T = T'$. Construction of $S'$ and $T'$ : First, $ABCD$ is trivially inscribed in a center with $AC$ as a diameter : denote its circumcenter by $O$. Moreover, let the midpoint of $HC$ be $M$. Since $O$ is the midpoint of $CA$ and $M$ is the midpoint of $CH$, $AH \parallel OM$ by Thales. but by definition $AH$ is perpendicular to $BD$, we have $OM \perp BD$ and, since $OB = OD$, this line is the perpendicular bisector of $BD$. Hence, $MB = MD$. We introduce the well defined circle $\Gamma$ with center $M$ passing through $B$ and $D$. Suppose that the ray $CH$ intersects $\Gamma$ at $E$. Let the tangent to $\Gamma$ through $E$ intersect $AB$, resp. $AD$ at $S'$, resp $T'$. Proof that $S'$ and $T'$ are valid : We will first prove $T'$ satisfies the angle condition. The same argument will hold for $S'$. Let $D' \neq D$ the second intersection of $\Gamma$ with $AD$ (if this line is tangent take $D' = D$), and denote $G$ as the midpoint of $D'D$. Note that $\angle CDA = 90^{\circ} = \angle MGD' = \angle MGA$ (because $M$ is the center of $\Gamma$ and $G$ is the midpoint of chord $GG'$) $\implies MG \parallel DC$. Thus, we can apply some sort of Thales on $HD', MG, CD$ : if $X$ is the intersection of $AD$ and $HC$, we know that (by parallelism and properties of midpoints) $$\frac{XG}{XM} = \frac{GD}{MC} = \frac{GD'}{MH} = \frac{GD'}{MH} \implies HD' \parallel MG \parallel DC.$$So, from the fact that $TECD$ is cyclic ($\angle TEC = \angle TEM = 90^{\circ} = \angle ADC = \angle TDC$), we immediately deduce that $TEHD'$ is cyclic as well.\newline We now prove the following : $TC \perp ED'$. By simple angle chasing, $$ \angle TED' + \angle CTE = \angle EDT + \angle CTE = \angle ECT + \angle CTE = 90^{\circ}$$by tangency and cyclicity, and we directly get the desired result. Finally, we can check that $$ \angle T'HC = 180^{\circ} - \angle EHT' = 180^{\circ} - \angle ED'T' = 90^{\circ} + (90^{\circ} - \angle ED'T') = 90^{\circ} + \angle D'T'C = 90^{\circ} + \angle DT'C$$by previously shown perpendicularity. That's exactly the result that we wanted. Now, it remains us to show that $(S'HT')$ is tangent to $BD$. To do this, we introduce $O', O_1$ and $O_2$ as the circumcenters of $S'HT', S'HC$ and $T'HC$ respectively, and $U, V$ on $AB$, resp. $AD$ such that $\angle S'CU = \angle T'CV = 90^{\circ}$. It's easy to check that $U \in (S'HC)$ and $V \in (T'HC)$ and consequently, $O_1$ resp. $O_2$ is the midpoint of $S'U$, resp. $T'V$. It follows that $$ O'O_1 \perp S'H \perp HU, O'O_2 \perp T'H \perp HV, O_1O_2 \perp HC = ME \perp S'T' \parallel UV$$by a bunch of well-known properties of the centers of circles (the last relation is due to the fact that $O_1O_2 \parallel S'T'$, so we can use Thales the same way as we did before since $S'O_1 = O_1U, T'O_2 = O_2V$). So the triangles $\triangle O'O_1O_2$ and $\triangle HUV$ have pairwise parallel sides ; by Desargues, $O'H, O_1U, O_2V$ must therefore concur. But $O_1U$ and $O_2V$ concur at $A \implies A, O', H$ are collinear. that means that $O'H \perp BD$, and we're done.

Let $(SHC) \cap AB = S' (\neq S)$ and $(THC) \cap AD = T' (\neq T)$. Claim : $\angle SHS'=\angle THT' = 90^{\circ}$ and $ST \parallel S'T'$. Proof : $\angle SS'C = 180 - \angle SHC = 90 - \angle S'SC \implies \angle SCS'=90^{\circ} \implies SHS'=90^{\circ}$. Note centre of $(SS'HC)$ is midpoint of $SS'$, which lies on $AB$. Similarly $\angle THT'=90^{\circ}$ and $(TT'HC)$ has centre on $AC$. Now perform $\sqrt{AB \cdot AD}$ inversion about $A$ followed by reflection along angle bisector of $\angle BAD$. Noting $H \leftrightarrow C$, observe that $(SS'HC)^*$ passes through $H,C$ and has centre along $AC$. Hence $(SS'HC)^* \equiv (TT'HC)$. Hence, $S \leftrightarrow T'$ and $S'\leftrightarrow T$. Hence $ST \parallel S'T'$ as claimed. $\square$ Now consider $X$ to be $H$-antipode in $(HST)$. We want to prove that $A$ lies on $HX$. $\angle XSH = 90^{\circ}=\angle SHS' \implies XS \parallel HS'$. Similarly $XT \parallel HT'$. As $ST \parallel S'T'$, indeed triangles $XST$ and $HS'T'$ are homothetic. Hence, $XH,SS', TT'$ concur. Hence $XH$ passes through $A$, and hence we are done!

Let $B'$ and $D'$ be the reflection of $C$ across $B$ and $D$, respectively. Observe that \[\angle THC = 90^\circ + \angle CTD =180^\circ - \angle TCD = 180^\circ -\angle TD'C\]so $D'CHT$ is cyclic. Similarly, $B'CHS$ is cyclic. We claim that $H$ and $C$ are isogonal conjugates with respect to $\triangle AST$. Clearly, $\angle HAB=\angle DAC$ because in $\triangle ABD$, $AH$ passes through the orthocenter and $AC$ passes through the circumcenter. Thus, it suffices to show that $\angle SHT+\angle SCT=180^\circ-\angle SAT$. This is true, since \begin{align*} \angle SHT+\angle SCT &= 360^\circ-\angle THC - \angle SHC + \angle SCT \\ &= \angle TD'C+\angle SB'C+\angle SCT \\ &= \angle TCD+\angle TCB+\angle SCT \\ &= \angle BCD \\ &= 180^\circ-\angle SAT \end{align*}as desired. Now, note that \[\angle ATH=180^\circ-\angle DTH=180^\circ-\angle D'TH+\angle D'TD=\angle DCH+90^\circ-\angle D'CT=90^\circ+\angle TCH\]while by conjugacy, \[\angle STH=\angle DTC=180^\circ-\angle ATH\]which implies that $\angle STH+\angle TCH=90^\circ$, or that $CH\perp ST$. Clearly, what this implies is that if $C'$ is the reflection of $C$ over $ST$ then $C'$ and $A$ are isogonal wrt $\triangle STH$, which implies that $HA$ passes through the circumcenter of $\triangle STH$ and we're done.

Infinityfun wrote: nima1376 wrote: $HS.HI=HC^{2}$ Why is this true? it's $HS\cdot HI=HC\cdot HM$

Reflect $C$ over $B$ and $D$ to get $B'$, $D'$. Then, as $\angle TD'C=\angle TCD=90 ^{\circ} -\angle DTC$ we find $CHTD'$ and $CHSB'$ are cyclic. Now, we invert the whole thing about $C$ and force overlay $B$ onto $D'$ and $D'$ onto $B$. Denote the inverse of a point with a prime, except $B,D$ and vice versa. The key claim is the following. Claim: $T'S' \parallel BD$ Proof. We start with some calculation. By inversion lengths, we find \[\frac{H'B}{H'C}=\frac{HD'}{CD'} \text{ and } \frac{H'C}{H'D}=\frac{CB'}{HB'}\]which when multiplied yield $\frac{H'B}{H'C}=\frac{HD'\cdot CB'}{CD'\cdot HB'}$. But notice that $AB'=AC=AD'$ so $A$ lies on the perpendicular bisector of $B'D'$, and as $AH\perp B'D'$ we get $H$ also does so $HB'=HD'$ implying \[\frac{H'B}{H'C}=\frac{CB'}{CD'}\cdot \frac{HD'}{HB'}=\frac{2BC}{2CD}=\frac{CB}{CD}\]Now, note that $T$ is an intersection of $H'B$ with $(A'C'B)$. But as $A$ inverts to the foot from $C$ to $B'D'$, circle $(A'C'B)$ is simply the circle with center $B$ through $C$. So we find $BT'=BC$ and similarly $DS'=DC$. So $\frac{BT'}{DS'}=\frac{CB}{CD}$ as well, implying the claim! $\blacksquare$ Now, it suffices to show that $(H'S'T')$ and $(H'B'CD')$ are tangent at $H'$. But by homothety at $H'$, we find $(H'S'T')$ is tangent to $(H'BD)$ at $H'$. So it suffices to show that $(H'B'CD')$ and $(H'BD)$ are tangent which is true after inverting back as $(H'BD)$ is sent to $(H'B'D')$ and this is clearly tangent to $BC$ since $HB'=HD'$ and $B'D' \parallel BD$. Remark: This problem is so good!

Consider the isogonal conjugate of $A$ in non-convex quadraliteral $CSHT$ (it exists because $\angle SAH=\angle TAC$), call this point $M$. By angles this point must lie on perpendiculars from $T$ and $S$ onto $HC$. One can easily check that $M$ can not be at infinity, so $TS \perp HC$. Call the second intersections of $(THC)$ and $(SHC)$ with lines $AD$ and $AB$ $F$ and $E$. Then $\angle SHE=\angle SCE=\angle THF=\angle TCF=90$, by Isogonal Lemma the intersection of $TE$ and $SF$, which is clearly not on $TS$, should be on isogonal of $HA$ in $\angle SHT$ and isogonal of $CA$ in $\angle SCT$. But if this two lines do not coincide, they have at most one point of intersections, which is $M$, but it lies on $ST$ - contradiction. So both isogonals are actually $CH$, which is perpendicular to $ST$, but then $AH$ passes through the circumcenter of $(SHT)$ and $BD \perp AH$, so we are done.

The point $S$ is uniquely determined given $A, B, C, D$, since $\angle CHS - \angle CSB$ decreases monotonically as $S$ is moved from $A$ to $B$. Let $E$ be the projection of $T$ onto $CH$. Let $S' = TE \cap AB$. Then $\angle HAS' = 90^\circ - \angle ABD = 90^\circ - \angle ACD = \angle CAT$ and $\angle HTS' = 90^\circ - \angle EHT = \angle THC - 90^\circ = \angle DTC$, so $H$ and $C$ are isogonal conjugates with respect to $\triangle AS'T$. Then $\angle CHS' = 180^\circ - \angle S'HE = 90^\circ + \angle TS'H = 90^\circ + \angle CS'B$, so $S'$ satisfies the angle condition and $S = S'$. Now let $C_1$ be the reflection of $C$ over $E$. Then \[ \angle C_1ST = \angle CST = 180^\circ - \angle AST - \angle CSB = 180^\circ - \angle AST - \angle TSH = 180^\circ - \angle ASH, \]so $SA$ and $SC_1$ are isogonal with respect to $\angle HST$. Similarly $TA$ and $TC_1$ are isogonal with respect to $\angle HTS$, and $A$ and $C_1$ are isogonal conjugates with respect to $\triangle HST$. To finish, we can calculate \[ \angle STH = 90^\circ - \angle C_1HT = 90^\circ - \angle SHA = \angle BHS, \]giving the desired tangency.

Let the circle centered at $A$ with radius $AC$ intersect $CB,CD$ at $P,Q$ respectively. Claim: $S,T$ lie on $(PCH)$ and $(QCH)$. Proof: \[\measuredangle SHC=\measuredangle BSC+90=\measuredangle PSB+90=180-\measuredangle BPS=180-\measuredangle CPS\]\[\measuredangle CHT=\measuredangle CTD+90=\measuredangle DTQ+90=180-\measuredangle TQD=180-\measuredangle TQC\]These yield $S\in (PCH)$ and $T\in (QCH)$.$\square$ New Problem Statement: $ABC$ is a triangle whose circumcenter is $O$. Let $E,F$ be the midpoints of $AC,AB$ respectively. $K$ is the altitude from $O$ to $EF$. $(AKB)$ and $(AKC)$ intersect the segments $OF,OE$ at $S,T$ respectively. Then, $(KST)$ is tangent to $EF$. Note that $OK$ passes through the midpoint of $BC$. Since $OF$ is the perpendicular bisector of $(AKB),$ $\overline{OSF}$ is the diameter of it. $S$ lies on the segment $OF$ hence $S$ is the midpoint of the arc $AKB$. Similarily, $T$ is the midpoint of the arc $AKC$. Take the inversion centered at $K$ with radius $KB=KC$. Let the angle bisectors of $\measuredangle BKA$ and $\measuredangle AKC$ intersect $A^*B$ and $A^*C$ at $M,N$ respectively. $S^*$ is on $A^*B$ and $T^*$ is on $A^*C$. \[\frac{S^*A}{S^*B}=\frac{A^*M}{MB}=\frac{KA^*}{KB}=\frac{KA^*}{KC}=\frac{A^*N}{NC}=\frac{T^*A}{T^*C}\]Thus, $S^*T^*\parallel BC\parallel \overline{KEF}$ as desired.$\blacksquare$

Don't really know why i haven't done this before but here we go i guess. Let $A', D'$ be reflections of $C$ over $A,D$ respectively, perpendiculars from $H$ to $SH,TH$ hit $BA,BD$ at $T',S'$ respectively and also let $A_1,D_1$ be reflections of $H$ over $BA,BD$ respectively. Now notice that from the condition+ reflection we get that $A'A_1SHCT'$ and $D'D_1THCS'$ are cyclic. Now consider $\sqrt{bd}$ inversion on $\triangle ABD$, it then holds that $H \to C, A' \to D_1, D' \to A_1$ which mean that $(SHC) \to (THC)$ therefore $S \to S'$ and $T \to T'$, finally let $O$ the circumcenter of $(STH)$. Claim: $H,C$ are isogonal conjugates in $\triangle BST$. Proof: let $H'$ be the isogonal conjugate of $H$ in $\triangle BST$ then by angle chasing: $$\angle SH'T=180-\angle HSA-\angle HTD=\angle ACD-\angle SHT=\angle ACD-\angle SA'C-\angle TD'C=\angle ACD-\angle SCA-\angle DCT=\angle SCT$$which means that $SH'CT$ is cyclic, but notice that in addition $B,H',C$ are colinear from isogonality, and since clearly $C,H'$ lies outside $\triangle BST$ we must have that $H'=C$ as desired. The finish: Now simply note that we want $B,O,H$ colinear but by angle chase: $$\angle THB=\angle TD_1B=\angle BT'A'=\angle SCA=90-\angle CSA=90-\angle TSH=\angle TOB$$And this is enough to conclude the colinearity, thus we are done .

skibi the snail Lemma: In a triangle $XYZ$, if the tangent to $(XYZ)$ at $X$ intersects $YZ$ at $P$, then $\frac{YP}{YZ}=\left(\frac{XY}{XZ}\right)^2$. Proof: i forgot similar triangles existed so lets use ratio lemma: \[\frac{YP}{YZ}=\frac{\sin\angle YXP}{\sin\angle ZXP}\times\frac{XY}{XZ}=\frac{XY}{XZ}\times \frac{\sin\angle XZY}{\sin\angle XYZ}=\left(\frac{XY}{XZ}\right)^2\]by alternate segment theorem and sine rule. Now we return to the original problem. Force overlay invert at $C$ that fixes $\triangle BCD$. Let the tangent at $C$ to $(ABC)$ intersect $BD$ at $K$. $KCHA$ is cyclic clearly. If $E$ is the foot from $C$ to $BD$, $C'$ the reflection of $C$ over perp bisector of $BD$, $H', E, C'$ collinear so $H'$ is the $C$-Why Point in $\triangle BCD$!!! Let $M,N$ be the midpoints of $BC,CD$ respectively. We claim $H'N\cap (CD)=S'$. But this is true because \[\angle CS'H'=180^\circ-\angle NS'C=180^\circ-\angle S'CD=90^\circ+\angle CDS'\]from reverse reconstruction and Thales yay! * Similarly we have $H'M\cap (BC)=T'$. (These are segments) By 2011 G4, $(MNH')$ is tangent to $(ABC)$. We want $(S'T'H')$ tangent to $(ABC)$. Clearly, it remains to prove $S'T'\parallel MN$. Now, \[\frac{MT'}{NS'}=\frac{BM}{DN}=\frac{BC}{CD}=\sqrt{\frac{BK}{DK}}=\sqrt{\frac{MF}{FN}}\]By Thales then by our lemma, where $F$ is the tangent at $C$ to $(ABC)$ intersecting $MN$, by homothety. Now suppose the tangent at $H'$ to $(ABC)$ intersects $MN$ at $F'$. Then, \[\frac{MH'}{NH'}=\sqrt{\frac{MF'}{F'N}}\]By our lemma. Consequently it only remains to prove $F'=F$. This is a consequence of radical axis theorem on $(H'MN),(ABC),(CMN)$. * Why are $S',T'$ unique one might ask... I'll just show it for $S'$. Note that \[90^\circ+\angle CDS'=180^\circ-\angle S'CD=180^\circ-\angle NS'C=\angle CS'H'\]and this gives the collinearity from the other direction.

I solved this in the contest. I found my solution written in an old note so I post it here. It is not particularly interesting, just straightforward bashing. Define a point $X$ on the ray $AB$ such that $\angle SCX=90$. We see that $\angle SXC=90-\angle XSC=180-\angle SHC$, so $X,C,H,S$ are on a circle. It follows that $\angle SHX=\angle SCX=90$. I will prove that $\frac {TD\cdot AB}{TH^2}=\frac {SB\cdot AD}{SH^2}$. This will be sufficient to solve the problem, because it would imply $$\frac {sinTHD}{sinSHB}=\frac {sinTDH}{sinSBH}\cdot\frac {SH}{SB}\cdot\frac {TD}{TH}=\frac {AB}{AD}\cdot\frac{SH}{SB}\cdot\frac{TD}{TH}=\frac{TH}{SH}=\frac{sinTSH}{sinSTH}$$Since $\angle THD+\angle SHB=\angle TSH+\angle STH$, we would get $\angle SHB=\angle STH$, which solves the problem. Now I will prove $\frac {TD\cdot AB}{TH^2}=\frac {SB\cdot AD}{SH^2}$. Denote $SB=x, \angle ABD=\alpha, \angle ADB=\beta, AD=b, AB=d$. We see that $SC^2=SB\cdot SX$, so $SX=\frac {SC^2}{x}=\frac{x^2+BC^2}{x}=\frac{x^2+(\frac{d}{tan\beta})^2}{x}$. Also, $SX=\frac{SH}{cosHSX}=\frac{SH}{\frac{SB^2+SH^2-HB^2}{2SB\cdot SH}}=\frac{2SB\cdot SH^2}{SB^2+SH^2-HB^2}$. We have $SH^2=SB^2+BH^2-2SB\cdot BHcos\alpha =x^2+(dcos\alpha)^2-2dxcos^2\alpha$, so we get $$SX=\frac{2SB\cdot SH^2}{SB^2+SH^2-HB^2}=\frac{2x(x^2+(dcos\alpha)^2-2dxcos^2\alpha)}{2x^2-2dxcos^2\alpha}=\frac{x^2+(dcos\alpha)^2-2dxcos^2\alpha}{x-dcos^2\alpha}$$Using the two equations on $SX$, we get $$\frac{x^2+(\frac{d}{tan\beta})^2}{x}=\frac{x^2+(dcos\alpha)^2-2dxcos^2\alpha}{x-dcos^2\alpha}$$Denote $cos^2\alpha =u, cos^2\beta =v$, and we can expand the above equation to get a quadratic equation on $x$. (note that $tan^2\beta =\frac{1-v}{v}$) The equation is $$u(1-v)x^2+(v-u+uv)dx-d^2uv=0$$This solves as $x=\frac{-(v-u+uv)\pm \sqrt{(v-u+uv)^2+4u^2v(1-v)}}{2u(1-v)}\cdot d$. The negative sign is apparently not possible, because $-(v-u+uv)-\sqrt{(v-u+uv)^2+4u^2v(1-v)}<-(v-u+uv)-|v-u+uv|\leq 0$. So we get $$\frac{x}{d}=\frac{-(v-u+uv)+\sqrt{Q}}{2u(1-v)}$$, where $Q=u^2+v^2+2u^2v+2uv^2-2uv-3u^2v^2$. Note that $Q$ is an expression symmetric in terms of $u$ and $v$. Now, we are ready to calculate $\frac {SB\cdot AD}{SH^2}$, which is what we need to do to solve the problem. Denote $k=\frac{-(v-u+uv)+\sqrt{Q}}{2u(1-v)}$. We saw that $\frac{x}{d}=k$, so $SH^2=x^2-2udx+ud^2=d^2(k^2-2uk+u)$. Therefore $\frac {SB\cdot AD}{SH^2}=\frac{xb}{d^2(k^2-2uk+u)}=\frac{kdb}{d^2(k^2-2uk+u)}$. Recall that $x$ was the root of the quadratic equation $u(1-v)x^2+(v-u+uv)dx-d^2uv=0$, so we get $u(1-v)k^2+(v-u+uv)k=uv$. We get $u=\frac {u(1-v)}{v}\cdot k^2+\frac{v-u+uv}{v}\cdot k$. Therefore, $k^2-2uk+u=k^2-2uk+\frac{u(1-v)}{v}\cdot k^2+\frac{v-u+uv}{v}\cdot k=\frac{u+v-uv}{v}\cdot k^2+\frac{v-u-uv}{v}\cdot k$. Therefore, $$\frac {SB\cdot AD}{SH^2}=\frac{kdb}{d^2(\frac{u+v-uv}{v}\cdot k^2+\frac{v-u-uv}{v}\cdot k)}=\frac{db}{d^2(\frac{u+v-uv}{v}\cdot k+\frac{v-u-uv}{v})}$$We can use $k=\frac{-(v-u+uv)+\sqrt{Q}}{2u(1-v)}$ to finish the calculation. $$(u+v-uv)k+v-u-uv=\frac{-u^2-v^2-2u^2v-2uv^2+2uv+3u^2v^2+(u+v-uv)\sqrt{Q}}{2u(1-v)}=\frac{-Q+(u+v-uv)\sqrt{Q}}{2u(1-v)}$$Denote $P=-Q+(u+v-uv)\sqrt{Q}$, which is also an expression symmetric in terms of $u$ and $v$. We get $$\frac {SB\cdot AD}{SH^2}=\frac{db}{d^2\cdot\frac{P}{2uv(1-v)}}=\frac{2uvdb}{P}\cdot\frac{1-v}{d^2}=\frac{2uvdb}{P}\cdot\frac{sin^2\beta}{d^2}=\frac{2uvdb}{P}\cdot\frac{1}{AC^2}$$Since $P$ is symmetric in terms of $u$ and $v$, similar arguments would yield $\frac {TD\cdot AB}{TH^2}=\frac{2uvdb}{P}\cdot\frac{1}{AC^2}$. Therefore, $\frac {TD\cdot AB}{TH^2}=\frac {SB\cdot AD}{SH^2}$, and we are done.

[asy][asy] unitsize(0.5cm); pair A, B, C, D, S, T, H, O1, O2; A=(3,3sqrt(5)); B=(0,0); D=(8,0); C=2*circumcenter(A,B,D)-A; H=foot(A,B,D); O1=extension(A,B,(C+H)/2,(C+H)/2+(C-H)*(0,1)); O2=extension(A,D,(C+H)/2,(C+H)/2+(C-H)*(0,1)); S=intersectionpoints(circle(O1,abs(O1-C)),A--B)[0]; T=intersectionpoints(circle(O2,abs(O2-C)),A--D)[0]; draw(A--B--D--A--2*O1-S--A--2*O2-T); draw(circumcircle(A,B,D)); draw(circle(O1,abs(O1-C))); draw(circle(O2,abs(O2-C))); draw(S--T); draw(O1--O2); draw(2*O1-S--2*O2-T); draw(T--H--S); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, S); label("$D$", D, E); label("$H$", H, SW); label("$S$", S, NNW); label("$T$", T, N); label("$S'$", 2*O1-S, SW); label("$T'$", 2*O2-T, SE); label("$O_1$", O1, NW); label("$O_2$", O2, NE); [/asy][/asy] Let $O_1$ be the circumcenter of $SCH$. Then, $\angle CHS=90^\circ+\angle CSO_1$, so $O_1$ lies on $AB$. Let the circumcircle of $SCH$ intersect $AB$ again at $S'$. Define $O_2$ and $T'$ similarly. Then, since $C$ and $H$ are $\sqrt{bc}$ inverses, $S,T$ and $T',S'$ are also $\sqrt{bc}$ inverses, so $ST\parallel S'T'\parallel O_1O_2\perp CH$. We have $\angle TSH=\angle SHC-90^\circ=\angle CSB=90^\circ-\frac12\angle CO_1S=90^\circ-\frac12\angle AO_1O_2-\frac12\angle O_2O_1H$. Similarly, $\angle STH=90^\circ-\angle AO_2O_1-\frac12O_1O_2H$. Subtracting gives $$\angle TSH-\angle STH=\frac12(\angle AO_2O_1-\angle AO_1O_2)+\frac12(\angle O_1O_2H-\angle O_2O_1H).$$ Since $O_1O_2\parallel ST$, we get $\frac{AO_1}{AO_2}=\frac{O_1H}{O_2H}$, so the angle bisectors of $\angle BAD$ and $\angle O_1HO_2$ Intersect on $O_1O_2$. Since the angle bisector of $\angle BAD$ is the angle bisector of $\angle CAH$, this intersection point is the midpoint of arc $CH$ in the circumcircle of $ACH$, so this makes an angle of $\frac{\angle HAC}2$ with $HC$. Therefore, $\angle HO_1O_2-\angle HO_2O_1=-\angle HAC$. Now, $\angle THD-\angle STH=\angle(ST,BD)=90^\circ-\angle DHC$, so $$\angle TSH-\angle THD=\frac12(\angle AO_2O_1-\angle AO_1O_2)+\frac12\angle HAC+\angle DHC-90^\circ.$$We have $\angle AO_2O_1-\angle AO_1O_2=-\angle HCD+\angle ADB+90^\circ-(-\angle BHC+90^\circ+\angle ABD)=-\angle DHC+\angle ADB+\angle BHC-\angle ABD$. Therefore, $$\angle TSH-\angle THD=-\frac12\angle DHC+\frac12(\angle ADB-\angle ABD)+\frac12\angle BHC+\frac12\angle HAC+\angle DHC-90^\circ=\frac12\angle DHC+\frac12\angle BHC-90^\circ=0.$$

First IMO P3 (>2010)