We have $KDSC$ is cyclic so $\angle KDP= \angle KCS=180^{\circ}- \angle KDS$. On the other hand, $\angle KCS= \angle KAP$ in $\Omega$. Hence, $\angle KAP= \angle KDP$. We obtain $PADK$ is cyclic. It follows that $\angle PKA= \angle PDA=90^{\circ}$. We also have $\angle PDA= \angle PMA=90^{\circ}$ then $PDMA$ is cyclic. But $PKDA$ is cyclic so $PKMA$ is cyclic. We get $\angle PKM= 180^{\circ}- \angle PAM= 180^{\circ}- \angle ABC= \angle AKC$. Thus, $\angle PKM= \angle AKC$ or $\angle PKA= \angle MKC=90^{\circ}$. $\blacksquare$ The problem still true for any triangle $ABC$.

Dear Mathlinkers, are you sure that KDSM is cyclic? Sincerely Jean-Louis

Sorry, my mistake. It is $KDSC$.

$\widehat{KDS}=\widehat{KCA}=\widehat{KAP}\Rightarrow$ $PDKMA$is cycle$\Rightarrow \widehat{AKM}=\widehat{APM}=90-\widehat{B}\Rightarrow \widehat{CKM}=90$ done

$CSKD$ cyclic. So, $\measuredangle KDP = \measuredangle KDS = \measuredangle KCS = \measuredangle KCA = \measuredangle KAP [\because PA$ touches $\Omega]$ So, $KPAD$ is cyclic and thus, $\measuredangle AKP = \measuredangle ADP = 90^0$ Now, using Zhao's Lemma No-1, we get $KP$ is the symmedian of $\triangle KAC$. So, $\measuredangle MKC = \measuredangle AKP = 90^0$ And so, obviously $\angle CKM = 90^0$ $[QED]$

First,notice PDMA is cyclic,now let circumcircles of PDMA and ABC intersect at T.We have <PAM=<ABC,so <PTM=<ATC and <TPM=<TAC and from this two we have <PMT=<ACT=<PDT from whic we have KCST is a cyclic,so K=T.Now,<CKM=180-<KCA-(90-<KMP)=90-(<KCA-<KCA)=90,so we are finished

Her is a bash(without calculations) approach: Let the feet of altitude from $M$ to $BC$ be $X$ and let $Y,Z$ be points on line $BS$ such that $\angle CYM=\angle CZM = 90{\circ}$. Now, all we have to do is prove that $(CSD),(ABC),(CYMZ)$ are co-axial circles for which we apply the Lemma in Mathematical Reflections Issue 5,2015 (see awesomemath.org) So, we need to prove that $\frac{SM}{SA}=\frac{DY.DZ}{DB.DT}$ where $T$ is the point at which the line $BS$ meets $(ABC)$ again. This is just(?) length chasing and fairly nasty but computationally feasible.

Double-posting because it's just so nice! Suppose that the points $A'$ and $C'$ are opposite to $A$ and $C$, respectively, in the circumcircle of triangle $ABC$. Clearly, points $C',M,K$ are collinear. Since $A'$ is symmetric to $C'$ in line $PM$ (perpendicular bisector of $AC$) and $C'K$ is a median; we have $A'K$ is a symmedian in triangle $AA'C$ and the points $P,A',K$ are collinear. Moreover, points $A,M,D,K,P$ lie on a circle with diameter $AP$. Therefore, $\angle MAK=\angle MKD$ and since we have the relations $\angle MSD=90^{\circ}-\angle MAK=90^{\circ}-\angle MKD=\angle DKC$, we conclude that the points $C,S,D,K$ lie on a circle. Comment. 800th post!

mathuz wrote: Let $M$ be the midpoint of the side $AC$ of acute-angled triangle $ABC$ with $AB>BC$. Let $\Omega $ be the circumcircle of $ ABC$. The tangents to $ \Omega $ at the points $A$ and $C$ meet at $P$, and $BP$ and $AC$ intersect at $S$. Let $AD$ be the altitude of the triangle $ABP$ and $\omega$ the circumcircle of the triangle $CSD$. Suppose $ \omega$ and $ \Omega $ intersect at $K\not= C$. Prove that $ \angle CKM=90^\circ $. V. Shmarov Claim: $AMDKP$ is cyclic. Proof: Firstly, $\angle AMP=\pi/2=\angle ADP$ and so $AMDP$ is cyclic. Call their circumcircle $\omega.$ Now, $\angle ASP+\angle SPA=\pi-\angle PAS=\pi-\angle B=\angle AKC.$ But $\angle AKC=\angle AKD+\angle DKC=\angle AKD+\angle PSA.$ Thus $\angle AKD=\angle SPA=\angle APS,$ and so $K$ also lies on $\omega.$ $\square$ Hence $\angle MKC=\angle MKD+\angle DKC=\angle SAD+\angle DSA=\pi/2,$ as desired. $\blacksquare$

Note $D\in (AP)$. It is easy to see that $PM\perp AC$, so $M\in (AP)$. So we know $MDAP$ cyclic. We have \begin{align*} \angle CKM &= \angle CKA-\angle MKA = 180-B - \angle MKA \\ &\implies \angle CKM=90 \text{ iff } \angle MKA = 90-B. \end{align*}But \[ \angle MPA = 90-\angle CAP = 90-\tfrac12(2B) = 90-B. \]So it suffices to show $K\in (AP)$ too, which is equivalent to $KDMA$ cyclic. Note \begin{align*} \angle AKD &= 180-B-\angle DKC = 180-B - (180-\angle DSC) = \angle DSC - B \\ &= 180-\angle BSC - B = \angle SBC + C-B. \end{align*}And \begin{align*} \angle AMD &= 180-\angle APD = 180-\angle APB = \angle PAB + \angle PBA \\ &= \angle PAC + A + B-\angle CBS = 2B+A-\angle CBS. \end{align*}So \[ \angle AKD + \angle AMD = C-B+2B+A = 180, \]and we conclude.

Let $K'$ be the point on $(ABC)$ such that $\angle AK'P=90^\circ$. Then angle chasing gives us \[\measuredangle K'CA=\measuredangle K'AP=\measuredangle K'DP\]and so $DSK'C$ is cyclic and $K'\equiv K$. Now notice that $PK$ is the $K$-symmedian in $\triangle CKA$, so \[\angle CKM=180^\circ-\angle AKP=90^\circ.\]

Same as the solutions above - posting for storage. Solved with L567. Let $\beta = \angle ABC$. Notice that $$\angle ADP = \angle PMA = 90^{\circ}$$meaning that $AMDP$ is cyclic. Therefore $$\angle SDM = \angle MAP = \angle CAP = \beta$$Reintroduce $K$ as the intersection of the circumcircles of $\triangle CSD$ and $AMDP$, we shall prove that $K$ also lies on the circumcircle of $\triangle ABC$. $$\angle CKA = \angle CKD + \angle DKA = \angle DSM + \angle DMS $$$$= 180^{\circ}-\angle SDM = 180^{\circ}-\beta = 180^{\circ}-\angle BCA$$Let $C'$ be the $C$ antipode. Then $\angle MKA = \angle MPA = 90^{\circ}-\beta = \angle C'KA$ meaning that $C,M,C'$ are collinear and in other words $\angle CKM = 90^{\circ}$ $\blacksquare$

We firstly have that $ADMP$ is cyclic . Now, $\angle PDK=\angle SDK=\angle SCK=\angle ACK=\angle PAK$. Hence $ADMKP$ is cyclic. Now,$PK$ is symmedian in $\triangle AKC$. Thus,$\angle MKC=180- \angle AKP=90$.

Nice problem, solved in a hotel room Let $X$ be the midpoint of $MP$, and redefine $K$ to be the intersection of $XC$ and $\Omega$. Then, a well known property of harmonics ($PML_1L_2$ is harmonic where $L_1$, $L_2$ are arc midpoints) will imply that $XM^2 = XL_1\cdot XL_2 = XK \cdot XC$ implying that $X$ lies on the radical axis of the point circle at $M$ and $\Omega$. Further, note that since $\angle CMP = 90^{\circ}$, $CM$ is diameter of $(CMK)$. So, all we need to show is that $K$ lies on $\omega$. For this, let $B'$ be the antipode of $B$. By projecting $(M,P;Y,\infty)$ at $C$ onto $\Omega$, we see that $P, K, B'$ are collinear. So, $\angle PKB = 90^{\circ}$. So, $B, D, M, K, P$ are concyclic. Thus, $\angle KDS = \angle KDP = \angle KBP = \angle KCB = \angle KCS$, so we are done. Edit: Went from 69 to 70 posts sad

[asy][asy] unitsize(4cm); import geometry; point B = dir(110); point C = dir(210); point A = dir(330); point O = (0,0); point M = intersectionpoint(perpendicular(O, line(A,C)), line(A,C)); point P = intersectionpoint(perpendicular(C, line(O,C)), perpendicular(A, line(O,A))); point S = intersectionpoint(line(B,P), line(A,C)); point D = intersectionpoint(perpendicular(A, line(B,P)), line(B,P)); point O_1 = circumcenter(triangle(C,S,D)); point K = reflect(line(O, O_1)) * C; dot("$A$", A, dir(0)); dot("$B$", B, dir(110)); dot("$C$", C, dir(180)); dot("$P$", P, dir(240)); dot("$D$", D, dir(180)); dot("$S$", S, dir(135)); dot("$M$", M, dir(45)); dot("$K$", K, dir(270)); draw(A--B--C--cycle); draw(P--A); draw(P--C); draw(circle(A,B,C), red); draw(B--P); draw(C--K); draw(K--M); draw(M--P); draw(circle(A,M,D), blue+dashed); [/asy][/asy] First, we prove the following: Claim: We have $AMDP$ is cyclic. Proof. Immediate from $\angle AMP = 90 = \angle ADP$. $\blacksquare$ Then, Claim: We have $(ADKP)$ is cyclic. Proof. Directing angles, \[ \measuredangle KDP = \measuredangle KDS = \measuredangle KCS = \measuredangle KCA = \measuredangle KAP \]where the last equality comes from $PA$ tangency to $\Omega$. $\blacksquare$ Then it follows that $AMDKP$ is cyclic. This implies that directing angles, \[ \measuredangle KCM = \measuredangle KCA = \measuredangle KAP = \measuredangle KMP = 90 - \measuredangle CMK \]so it follows that $\measuredangle CKM = 90$ as needed.