[asy][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10.5cm); 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 = -2.5, xmax = 4.5, ymin = -1, ymax = 5; /* image dimensions */ /* draw figures */ draw((-1.06,4.28)--(-2.19,-0.53)); draw((-2.19,-0.53)--(4,-0.54)); draw((-1.06,4.28)--(4,-0.54)); draw((-1.62,1.88)--(0.52,2.78)); draw((-1.62,1.88)--(4,-0.54)); draw((-1.62,1.88)--(-0.58,-0.53)); draw((-0.58,1.63)--(0.52,2.78)); draw((-0.58,1.63)--(-0.58,-0.53)); draw((-1.62,1.88)--(-0.58,1.63)); /* dots and labels */ dot((-1.06,4.28),dotstyle); label("$A$", (-1,4.38), NE * labelscalefactor); dot((-2.19,-0.53),dotstyle); label("$B$", (-2.12,-0.43), NE * labelscalefactor); dot((4,-0.54),dotstyle); label("$C$", (4.06,-0.44), NE * labelscalefactor); dot((-1.62,1.88),dotstyle); label("$K$", (-1.56,1.97), NE * labelscalefactor); dot((0.52,2.78),dotstyle); label("$L$", (0.59,2.88), NE * labelscalefactor); dot((-0.58,-0.53),dotstyle); label("$M$", (-0.52,-0.43), NE * labelscalefactor); dot((-0.58,1.63),dotstyle); label("$O$", (-0.5,1.72), NW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $O$ be the intersection of the perpendicular to the sides $AB$ and $AC$ passing through $K$ and $L$, respectively. Simply angle chasing, we have $\angle BMK = \angle ALK = \angle AOK =\angle BOK$, and thus $BMOK$ is cyclic. Therefore, $\angle BMO = 90^\circ$ and so the perpendiculars are concurrent.

We can judge it little bit reverse as well: take $O$ on the perpendicular bisector of $AB$ so that $\angle KOA=\angle KLA\iff\angle KOB=\angle KMB$, or $AKOL, KOMB$ are cyclic $\iff OM\bot BC, OL\bot AC$, done. Best regards, sunken rock

Some easier problem. We $ AK=BK $ and $ \angle ALK=\angle BMK $ so circumradius of $ ALK $ and $ BMK $ are equal $ (*) $.Let $ \angle PMB=\angle PMK=90 $ and $ \angle QLA= \angle QKA=90 $, then $ P,Q $ are lies on perpendicular bisector of $ AB $ and by $ (*) $ we have $ AQ=BP $ so $ P=Q $, so problem is proved. ________ I'm sorry for my bad english.

lol Motivated by $AK=KB$, we will try to use Carnot's Theorem. It suffices to show that $BM^2 - CM^2 = AL^2- CL^2$. Denote $\angle ALK = \angle KMB = \alpha$, we have $BM= \frac{c}{2 \sin \alpha} \cdot \sin (B+\alpha)$ and $AL= \frac{c}{2 \sin \alpha} \cdot \sin (C+\alpha)$. Now plugging it all in, it suffices to show that $\frac{ac \sin (B+\alpha)}{\sin \alpha} - a^2 = \frac{bc \sin (A+ \alpha)}{\sin \alpha} - b^2$. Using trig sum formulas and LoC, it is easy to see that this holds. $\blacksquare$

Carnot theorem:$$CM^2+AL^2=CL^2+BM^2$$Remainig easy by sinus.