Let $A = \angle CAB, B = \angle ABC, C = \angle BCA = 90^{\circ}$, $a=BC$, $b=CA$ and $c= AB = \sqrt{a^2+b^2}$, Part (a)Without loss of generality assume that $AC \geq BC$. Since, $MP \equiv MK \parallel BC$, $\angle PMC = 180^{\circ} - \angle MCB = 90^{\circ}$ and $\angle MPC = \angle PCB = \angle NCB = \angle NCA = \angle PCM$. $\therefore MP = CP = \frac12b$. Thus, $CP = \sqrt{\left(\frac{b}2\right)^2+\left(\frac{b}2\right)^2} = \frac{b}{\sqrt2}$. Similarly, $CQ = \frac{a}{\sqrt2}$. By Power of a Point Theorem, \[\frac{ab}2 = CP\cdot CQ = CS \cdot CA = CS\cdot b \implies CS = \frac{a}2 = CL. \text{ Similarly }CT = \frac{b}2 = CM.\]\[MS = MC - CS = \frac{b-a}2 = CT - CL = LT\]Let $X = MT \cap CN$, $MX = TX \because \triangle MCT$ is isosceles with $MC = CT$. \[\text{Therefore, } \frac{MS}{CS}\cdot\frac{CL}{TL}\cdot\frac{TX}{MX} = \frac{MS}{TL}\cdot \frac{LC}{CS}\cdot\frac{TX}{MX} = 1\]Thus, by Ceva's Theorem in $\triangle MCT$, $ML$, $ST$ and $CX=CN$ are concurrent. [asy][asy] import geometry; import olympiad; unitsize(2.25cm); pair B = (2.3,0), C = (0,0), A = (0, 3); pair K = (A+B)/2, L = (B+C)/2, M = (C+A)/2; pair N = (length(C-A)*B+length(B-C)*A)/(length(B-C)+length(C-A)); pair P = intersectionpoint(line(M,K), line(C,N)), Q = intersectionpoint(line(L,K), line(C,N)); path u = circumcircle(A,P,Q), v = circumcircle(B,P,Q); pair S = intersectionpoints(u,A--C)[1]; pair T = intersectionpoints(v,B--C)[0]; pair X = intersectionpoint(C--N, M--T); real[] s = {5,7}, r = {6,8}; draw(anglemark(B,C,N,t=s.length ... s)); draw(anglemark(N,C,A,t=r.length ... r)); draw(A--B--C--cycle); draw(C--N); draw(K--M); draw(K--L); draw(N--P, dashed); draw(K--P, dashed); draw(u,dotted); draw(v,dotted); draw(S--T); draw(M--L); draw(M--T,dashed); dot(A); dot(B); dot(C); dot(K); dot(L); dot(M); dot(N); dot(P); dot(Q); dot(S); dot(T); dot(X); label("$A$", A, dir(135)); label("$B$", B, dir(-45)); label("$C$", C, dir(225)); label("$K$", K, dir(45)); label("$L$", L, dir(-90)); label("$M$", M, dir(180)); label("$N$", N, dir(0)); label("$P$", P, 2dir(45)); label("$Q$", Q, dir(-30)); label("$S$", S, dir(225)); label("$T$", T, dir(-90)); label("$X$", X, dir(0)); [/asy][/asy] Part (b)Since, $MP \equiv MK \parallel BC$, $\angle MPC = \angle PCB = \angle NCB = 90^{\circ} - B = A$. $\angle MCP = \angle ACN = 90^{\circ} - A = B$. Therefore, $\triangle PMC \sim \triangle ACB$. \[\frac{\frac12b}{a} = \frac{MC}{BC} = \frac{CP}{AB} = \frac{CP}c \implies CP = \frac{bc}{2a}.\text{ Similarly, } CQ = \frac{ca}{2b}\] By Power of a Point theorem, \[\frac{c^2}4 = CQ \cdot CP = CS \cdot CA = CS\cdot b \implies CS = \frac{c^2}{4b} = \frac14\left(\frac{a^2}b+b\right)\] Similarly, $\displaystyle CT = \frac{c^2}{4a}= \frac14\left(a+\frac{b^2}a\right)$ Let $C$ have the cartesian co-ordinates $(0,0)$, $A = (0, b)$ and $B = (0,a)$. Then $M = \left(0, \frac12b\right)$, $L = \left(\frac12a, 0\right)$. As proved above $CS = \frac14\left(\frac{a^2}b+b\right)$, therefore $S = \left(\frac14\left(\frac{a^2}b+b\right),0\right)$ and similarly $CT = \left(0,\frac14\left(a+\frac{b^2}a\right)\right)$. Let $X$ be the midpoint of $ML$, $X = \frac12(M+L) = \left(\frac14a,\frac14b\right)$. \begin{align*} 32\;\text{ar}(\triangle STX) &= 16\begin{vmatrix} 0 &\frac14\left(\frac{a^2}b+b\right) & 1 \\ \frac14\left(a+\frac{b^2}a\right) & 0 & 1 \\ \frac14a & \frac14b & 1 \end{vmatrix} = \begin{vmatrix} 0 & \frac{a^2}b+b & 1 \\ a+\frac{b^2}a& 0 & 1 \\ a & b & 1 \end{vmatrix} \\ &= \begin{vmatrix} -a & \frac{a^2}b & 0 \\ \frac{b^2}a& -b & 0 \\ a & b & 1 \end{vmatrix} = \begin{vmatrix} -a & \frac{a^2}b \\ \frac{b^2}a& -b \end{vmatrix} = (-a)(-b)-\frac{a^2}b \cdot \frac{b^2}a = ab - ab = 0 \end{align*}[asy][asy] import geometry; import olympiad; unitsize(2.25cm); pair B = (2.3,0), C = (0,0), A = (0, 3); pair K = (A+B)/2, L = (B+C)/2, M = (C+A)/2; pair N = foot(C,A,B); pair P = intersectionpoint(line(M,K), line(C,N)), Q = intersectionpoint(line(L,K), line(C,N)); path u = circumcircle(A,P,Q), v = circumcircle(B,P,Q); pair S = intersectionpoints(u,A--C)[1]; pair T = intersectionpoints(v,B--C)[0]; pair X = (M+L)/2; draw(rightanglemark(C,N,B,3)); draw(rightanglemark(A,C,B,4)); draw(A--B--C--cycle); draw(C--N); draw(K--M); draw(K--L); draw(N--P, dashed); draw(K--P, dashed); draw(u,dotted); draw(v,dotted); draw(S--T); draw(M--L); dot(A); dot(B); dot(C); dot(K); dot(L); dot(M); dot(N); dot(P); dot(Q); dot(S); dot(L); dot(X); dot(T); label("$A$", A, dir(135)); label("$B$", B, dir(-45)); label("$C$", C, dir(225)); label("$K$", K, dir(45)); label("$L$", L, dir(-90)); label("$M$", M, dir(180)); label("$N$", N, dir(0)); label("$P$", P, dir(-45)); label("$Q$", Q, dir(-30)); label("$S$", S, dir(225)); label("$T$", T, dir(-90)); label("$X$", X, dir(225)); [/asy][/asy]

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.22, xmax = 15.18, ymin = -5.24, ymax = 8.28; /* image dimensions */ pair A = (0.,5.), B = (9.,0.), C = (0.,0.), M = (0.,2.5), K = (4.5,2.5), L = (4.5,0.), Q = (4.5,4.5), P = (2.5,2.5), T = (2.5,0.), X = (1.6071428571428572,1.6071428571428572); draw(A--B--C--cycle, linewidth(1.)); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--A, linewidth(1.)); draw(M--K, linewidth(1.)); draw(L--Q, linewidth(1.)); draw(C--Q, linewidth(1.)); draw(P--T, linewidth(1.)); draw((0.,4.5)--Q, linewidth(1.)); draw((0.,4.5)--T, linewidth(1.)); draw(M--L, linewidth(1.)); draw((0.,4.5)--P, linewidth(1.)); draw(Q--B, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (0.08,5.2), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (9.14,-0.38), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (-0.36,-0.36), NE * labelscalefactor); dot(M,linewidth(4.pt) + dotstyle); label("$M$", (-0.36,2.5), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (4.58,2.64), NE * labelscalefactor); dot(L,linewidth(4.pt) + dotstyle); label("$L$", (4.5,-0.4), NE * labelscalefactor); dot(Q,linewidth(4.pt) + dotstyle); label("$Q$", (4.58,4.66), NE * labelscalefactor); dot(P,linewidth(4.pt) + dotstyle); label("$P$", (2.6,2.14), NE * labelscalefactor); dot(T,linewidth(4.pt) + dotstyle); label("$T$", (2.24,-0.46), NE * labelscalefactor); dot((0.,4.5),linewidth(4.pt) + dotstyle); label("$S$", (-0.34,4.28), NE * labelscalefactor); dot(X,linewidth(4.pt) + dotstyle); label("$X$", (1.44,1.14), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: $SQLC,TPMC$ is square. Proof: As $P$ lies on $MK$, $PC=PA$. $\angle APQ = \angle ACP+\angle CAP=2\cdot \angle ACP=90^{\circ}$. So $SQ\perp AC$. Also, $QC$ bisects $\angle SCL$, so $SQLC$ is square. Similarly $MPTC$ is square. Main problem: Let $X=ML\cap ST$. Consider $r$ to be the reflection about $CN$. From the claim, $r$ maps $L$ to $S$ and $T$ to $M$. So, $r$ maps $ML$ to $TS$ and $TS$ to $ML$. Suppose $X'$ is the reflection of $X$ about $CN$. So, $r$ takes $X=ML\cap TS$ to $X'=TS\cap ML=X$. So, $X=X'\Rightarrow X$ lies on $CN$. So, $ML,ST,CN$ concur.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.78, xmax = 15.62, ymin = -3.5, ymax = 10.02; /* image dimensions */ pair A = (0.,5.), B = (7.,0.), C = (0.,0.), L = (3.5,0.), M = (0.,2.5), K = (3.5,2.5), Q = (3.5,4.9), P = (1.7857142857142856,2.5), T = (2.6428571428571437,0.); draw(A--B--C--cycle, linewidth(1.)); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--A, linewidth(1.)); draw(M--K, linewidth(1.)); draw(Q--L, linewidth(1.)); draw(Q--C, linewidth(1.)); draw(P--T, linewidth(1.)); draw((2.3648648648648645,3.3108108108108105)--L, linewidth(1.)); draw(Q--(0.,3.7), linewidth(1.)); draw(M--(2.3648648648648645,3.3108108108108105), linewidth(1.)); draw(Q--B, linewidth(1.)); draw(B--P, linewidth(1.)); draw(P--A, linewidth(1.)); draw(A--Q, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (-0.3,5.14), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (7.1,-0.32), NE * labelscalefactor); dot(C,linewidth(4.pt) + dotstyle); label("$C$", (0.08,-0.36), NE * labelscalefactor); dot(L,linewidth(4.pt) + dotstyle); label("$L$", (3.52,-0.36), NE * labelscalefactor); dot(M,linewidth(4.pt) + dotstyle); label("$M$", (-0.44,2.4), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (3.58,2.66), NE * labelscalefactor); dot(Q,linewidth(4.pt) + dotstyle); label("$Q$", (3.64,5.06), NE * labelscalefactor); dot((2.3648648648648645,3.3108108108108105),linewidth(4.pt) + dotstyle); label("$N$", (2.54,3.2), NE * labelscalefactor); dot(P,linewidth(4.pt) + dotstyle); label("$P$", (1.6,1.9), NE * labelscalefactor); dot((0.,3.7),linewidth(4.pt) + dotstyle); label("$S$", (-0.32,3.54), NE * labelscalefactor); dot(T,linewidth(4.pt) + dotstyle); label("$T$", (2.5,-0.36), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] If $AC=BC$, it is same as part (a). So, assume $AC\neq BC$. Claim: $PT||NL$ and $QS||NM$. Proof: As $Q$ lies on $KL$, $QC=QB\Rightarrow \angle CPT=\angle CBQ=\angle QCB=90^{\circ}-\angle B$. As $L$ is the midpoint of the hypotenuse $BC$ of triangle $BNC$, $LN=LC$. So, $\angle CNL=90^{\circ}-\angle B$. Proved. Main Problem: We'll proceed by coordinate geometry. Let $C=(0,0),A=(0,a),B=(b,0)$. So, $L=\left(\frac{b}{2},0\right),M=\left(0,\frac{a}{2}\right)$. As $CN\perp AB$, slope of $CN$ is $\frac{b}{a}$. So, the equation of $CN$ is $y=\frac{b}{a}x$. We know the equation of line $AB: y=\frac{-a}{b}x+a$. Equating the equations we get, $N=\left(\frac{a^2b}{a^2+b^2},\frac{ab^2}{a^2+b^2}\right)$. Equation of $CN$ is $y=\frac{b}{a}x$. So, $Q=\left(\frac{b}{2},\frac{b^2}{2a}\right),P=\left(\frac{a^2}{2b},\frac{a}{2}\right)$. Let $T=(x_0,0)$. From the claim $TP||NL$. The slope of $NL$ is $\frac{\frac{b^2a}{a^2+b^2}}{\frac{a^2b}{a^2+b^2}-\frac{b}{2}}=\frac{2b^2a}{2a^2-a^2b-b^3}=\frac{2ab}{a^2-b^2}$. Slope of $PT=\frac{\frac{a}{2}}{\frac{a^2}{2b}-x_0}.$ So, $x_0=\frac{a^2+b^2}{4b}$. So, $T=\left(\frac{a^2+b^2}{4b},0\right), S=\left(0,\frac{a^2+b^2}{4a}\right)$. Let $X=ML\cap ST$. We have the equations of lines $ML$ and $ST$ as $y=\frac{-a}{b}x+\frac{a}{2}$ and $y=\frac{-b}{a}x+\frac{a^2+b^2}{4a}$ respectively. Equating, we get \[x=\frac{\frac{b^2-a^2}{4a}}{\frac{b^2-a^2}{ab}}=\frac{b}{4}.\]So, $y=\frac{a}{4}$ which gives $X$ as the midpoint of $ML$. Proved.