Let $\frac{KD}{MD} = \frac{LE}{NE}$ hold. Using the power of points D and E, wrt to $\odot AKC$ and $\odot ALB$, respectively, we get: $KD^2 \cdot AE \cdot EB = LE^2 \cdot AD \cdot DC$ Let $\odot PBC$ intersect $AB$ and $AC$ again at $E'$ and $D'$. Now using the power of points $D$ and $E$ wrt $\odot PBC$ we get that: $KD^2 \cdot AE \cdot EB = LE^2 \cdot AD \cdot DC \implies DD' \cdot CD \cdot AE \cdot EB = EE' \cdot EB \cdot AD \cdot DC \implies DD' \cdot AE = EE' \cdot AD$. From this is fairly easy to conlcude that $E'D' \parallel ED$ and hence $B,D,E,C$ are concyclic points. Now let $F$ be the intersection of $AB$ and $EF$. Then $AP$ is the polar line of $F$ wrt $\odot BEDC$. But the perpedicular from a point to it's polar line passes through the center of the circle, hence the center of $\odot BEDC$ lies on $BC$ and it follows that it's the midpoint of $BC$, hence $BD$ and $CE$ are altitudes in $\triangle ABC$, hence $P$ is the orthocenter of $\triangle ABC$. For the other direction using power of point multiple times we get: $$\frac{\frac{MD}{KD}}{\frac{NE}{LE}} = \frac{AD\cdot DC \cdot EL^2}{AE \cdot EB \cdot KD^2} = \frac{AD\cdot DC \cdot EP \cdot EC}{AE \cdot EB \cdot DP \cdot DB}$$ Now from $\triangle DEP \sim \triangle BCP$ we get $\frac{EP}{DP} = \frac{PB}{PC}$, substituting this and using $\frac{DC}{EB} = \frac{PC}{PB}$ and $\frac{AD}{AE} = \frac{DB}{EC}$ from $\triangle PDC \sim \triangle PEB$ and $\triangle AEC \sim \triangle ADB$, respectively we get that the ratio is equal to $1$ Hence $\frac{KD}{MD} = \frac{LE}{NE}$

Hello. My solution. We will use Power of Point. $\frac{KD}{MD}=\frac{KD^2}{KD\cdot MD}=\frac{DP\cdot DB}{AD\cdot CD}$.Similarly $\frac{EP\cdot EC}{AE\cdot BE}$. Hence,from the given relation,we obtain $\frac{DP\cdot DB}{AD\cdot CD}=\frac{EP\cdot EC}{AE\cdot BE}\Leftrightarrow \frac{AD}{DB}\cdot \frac{CD}{DP}=\frac{BE}{EP}\cdot \frac{AE}{CE}$. We apply the sine law in $\triangle{ADB},\triangle{CDP},\triangle{BEP},\triangle{AEC}$ and the above relation is written as $\frac{\sin \hat{ABD}}{\sin \hat{DAB}}\cdot \frac{\sin \hat{CPD}}{\sin \hat{DCP}}=\frac{\sin \hat{BPE}}{\sin \hat{EBP}}\cdot \frac{\sin \hat{ACE}}{\sin \hat{CAE}}$. Since $\angle{EBP}=\angle{ABD},\angle{EBP}=\angle{CPD},\angle{CAE}=\angle{DAP},\angle{DCP}=\angle{ACE}$,the above relation is equivalent to $\sin ^2 \hat{ABD}=\sin ^2\hat{ACE}$.We have to show that this is true if and only if $P$ is the orthocenter of $\triangle{ABC}$. If $P$ is the orthocentre of $\triangle{ABC}$ the above relation is obviously true since $BCDE$ is cyclic. If the above relation is true then,since $\angle{ABD},\angle{ACE}\in (0,\pi)$,we obtain $\angle{ABD}=\angle{ACE}$. This imples that $BCDE$ is cyclic whence it follows that $H$ is the orthocentre of $\triangle{ABC}$. [asy][asy]import graph; size(10cm); 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 = -6.611738766406639, xmax = 18.660588160486476, ymin = -7.225153289872827, ymax = 7.073663260869347; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((2.0015509204177233,3.7394338415460675)--(0.,0.)--(6.,0.)--cycle, aqaqaq); /* draw figures */ draw((2.0015509204177233,3.7394338415460675)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(2.0015509204177233,3.7394338415460675), uququq); draw(circle((3.,-0.7996218509572753), 3.104737525867258)); draw((1.336176154367909,2.4963353562183985)--(6.,0.)); draw((2.7993828472505475,2.993284610260559)--(0.,0.)); draw(circle((1.2562350762982029,1.7329808662312116), 2.140408664167998)); draw(circle((3.82850451002141,1.6855134687523432), 2.7488812845124677)); draw((-0.6510179986258853,2.7044410856113177)--(3.3233703073617056,2.288229626825479)); draw((1.079995892317005,1.640250959459031)--(4.518769802184089,4.346318261062086)); /* dots and labels */ dot((2.0015509204177233,3.7394338415460675),linewidth(3.pt) + dotstyle); label("$A$", (1.9628007266463823,3.962126768930436), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.4599223434129202,-0.3370190710767628), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.285698753614942,-0.17075376234720263), NE * labelscalefactor); dot((2.0015509204177233,2.1401901396315184),linewidth(3.pt) + dotstyle); label("$P$", (1.9152963527236508,1.5631558858324963), NE * labelscalefactor); dot((2.7993828472505475,2.993284610260559),linewidth(3.pt) + dotstyle); label("$D$", (2.960392579023742,3.3445699079349267), NE * labelscalefactor); dot((1.336176154367909,2.4963353562183985),linewidth(3.pt) + dotstyle); label("$E$", (0.8464479394621939,2.845773981746246), NE * labelscalefactor); dot((3.3233703073617056,2.288229626825479),linewidth(3.pt) + dotstyle); label("$L$", (3.530445066096519,1.8006777554461537), NE * labelscalefactor); dot((1.079995892317005,1.640250959459031),linewidth(3.pt) + dotstyle); label("$K$", (0.5851738828871711,1.6581646336779592), NE * labelscalefactor); dot((4.518769802184089,4.346318261062086),linewidth(3.pt) + dotstyle); label("$M$", (4.623045666319342,4.4846748820804825), NE * labelscalefactor); dot((-0.6510179986258853,2.7044410856113177),linewidth(3.pt) + dotstyle); label("$N$", (-0.9587182696016001,2.9645349165530748), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Edit:Too late...

Lemma: Let K a point on the $A-$ altitude if $D,E$ are the respective intersections of $BK,CK$ with $CA,BA$ and $BCDE$ are cyclic then K is the orthocenter proof : $A',B',C'$ the feet of $A,B,C$ resp. it 's clear that $B'C' \parallel DE$ but if $S,T$ are resp. the points where $B'C',DE$ hit $BC$ then $(B,C;A',S)=-1$ and $(B,C;A',T)=-1$ so T=S then $B'C'\equiv DE$ thus $K$ is the orthocenter . Back to the problem $\frac{KD}{MD}=\frac{LE}{NE}\implies \dfrac{KD^2}{MD\cdot KD}=\dfrac{LE^2}{NE\cdot LE}$$\implies\dfrac{KD^2}{DA\cdot DC}=\dfrac{LE^2}{EA\cdot EB} $ that's means $D,E$ are on the coaxal circle with $(ABC)$ and $(BCP)$ so $ DEBC$ are cyclic. for the converse: if $P$ is the orthocenter then since $DEBC$ is cyclic then coaxal with $(ABC)$ and $(BCH)$ thus $\dfrac{DK^2}{DA\cdot DC}=\dfrac{EL^2}{EA\cdot EB} \implies\dfrac{DK^2}{DK\cdot DM}=\dfrac{EL^2}{EL\cdot EN} $ therefore $\dfrac{DK}{ DM}=\dfrac{EL}{ EN}.$ R HAS

Let $(BPC) \cap AB = {B, X}$ , $(BPC) \cap AC = {C,Z}$ Note that $\frac{EL}{EN}=\frac{EL^2}{EA . EB}$ and $\frac{DK}{DM}=\frac{DK^2}{DA . DC}$ . Equality $\iff \frac{EL^2}{DK^2}=\frac{EA . EB}{DA . DC} \iff \frac{EX.EB}{DZ.DC}=\frac{EA . EB}{DA . DC} \iff \frac{AE}{EX}=\frac{AD}{DZ}\iff ED||XZ \iff BEDC$ concyclic. Now , it is not difficult to conclude .

PROF65 wrote: Lemma: Let K a point on the $A-$ altitude if $D,E$ are the respective intersections of $BK,CK$ with $CA,BA$ and $BCDE$ are cyclic then K is the orthocenter proof : $A',B',C'$ the feet of $A,B,C$ resp. it 's clear that $B'C' \parallel DE$ but if $S,T$ are resp. the points where $B'C',DE$ hit $BC$ then $(B,C;A',S)=-1$ and $(B,C;A',T)=-1$ so T=S then $B'C'\equiv DE$ thus $K$ is the orthocenter . Back to the problem $\frac{KD}{MD}=\frac{LE}{NE}\implies \dfrac{KD^2}{MD\cdot KD}=\dfrac{LE^2}{NE\cdot LE}$$\implies\dfrac{KD^2}{DA\cdot DC}=\dfrac{LE^2}{EA\cdot EB} $ that's means $D,E$ are on the coaxal circle with $(ABC)$ and $(BCP)$ so $ DEBC$ are cyclic. for the converse: if $P$ is the orthocenter then since $DEBC$ is cyclic then coaxal with $(ABC)$ and $(BCH)$ thus $\dfrac{DK^2}{DA\cdot DC}=\dfrac{EL^2}{EA\cdot EB} \implies\dfrac{DK^2}{DK\cdot DM}=\dfrac{EL^2}{EL\cdot EN} $ therefore $\dfrac{DK}{ DM}=\dfrac{EL}{ EN}.$ R HAS Your lemma is wrong if $ABC$ is isosceles. Eray wrote: In an acute triangle $ABC$, a point $P$ is taken on the $A$-altitude. Lines $BP$ and $CP$ intersect the sides $AC$ and $AB$ at points $D$ and $E$, respectively. Tangents drawn from points $D$ and $E$ to the circumcircle of triangle $BPC$ are tangent to it at points $K$ and $L$, respectively, which are in the interior of triangle $ABC$. Line $KD$ intersects the circumcircle of triangle $AKC$ at point $M$ for the second time, and line $LE$ intersects the circumcircle of triangle $ALB$ at point $N$ for the second time. Prove that\[ \frac{KD}{MD}=\frac{LE}{NE} \iff \text{Point P is the orthocenter of triangle ABC}\] So in problem we must have $ABC$ isn't isosceles

Well, someone can check my solution? It's clearly: $P_{D/(ABC)}=P_{D/(AKC)}=\overline{DK}.\overline{DM}$ and $P_{D/(BPC)}=DK^2$, hence: $\frac{P_{D/(ABC)}}{P_{D/(BPC)}}=\frac{\overline{DM}}{\overline{DK}}$. Likewise, $\frac{P_{E/(ABC)}}{P_{E/(BPC)}}=\frac{\overline{EN}}{\overline{EL}}$, hence, $\frac{\overline{DM}}{\overline{DK}}=\frac{\overline{EN}}{\overline{EL}}$ iff $\frac{P_{D/(ABC)}}{P_{D/(BPC)}}=\frac{P_{E/(ABC)}}{P_{E/(BPC)}}$, which is actually equals to $BEDC$ is concyclic and yet it is enough

PROF65 wrote: Lemma: Let K a point on the $A-$ altitude if $D,E$ are the respective intersections of $BK,CK$ with $CA,BA$ and $BCDE$ are cyclic then K is the orthocenter proof : $A',B',C'$ the feet of $A,B,C$ resp. it 's clear that $B'C' \parallel DE$ but if $S,T$ are resp. the points where $B'C',DE$ hit $BC$ then $(B,C;A',S)=-1$ and $(B,C;A',T)=-1$ so T=S then $B'C'\equiv DE$ thus $K$ is the orthocenter Your lemma can be proven by using direct Brocard's Theorem

tenplusten wrote: PROF65 wrote: Lemma: Let K a point on the $A-$ altitude if $D,E$ are the respective intersections of $BK,CK$ with $CA,BA$ and $BCDE$ are cyclic then K is the orthocenter proof : $A',B',C'$ the feet of $A,B,C$ resp. it 's clear that $B'C' \parallel DE$ but if $S,T$ are resp. the points where $B'C',DE$ hit $BC$ then $(B,C;A',S)=-1$ and $(B,C;A',T)=-1$ so T=S then $B'C'\equiv DE$ thus $K$ is the orthocenter . Back to the problem $\frac{KD}{MD}=\frac{LE}{NE}\implies \dfrac{KD^2}{MD\cdot KD}=\dfrac{LE^2}{NE\cdot LE}$$\implies\dfrac{KD^2}{DA\cdot DC}=\dfrac{LE^2}{EA\cdot EB} $ that's means $D,E$ are on the coaxal circle with $(ABC)$ and $(BCP)$ so $ DEBC$ are cyclic. for the converse: if $P$ is the orthocenter then since $DEBC$ is cyclic then coaxal with $(ABC)$ and $(BCH)$ thus $\dfrac{DK^2}{DA\cdot DC}=\dfrac{EL^2}{EA\cdot EB} \implies\dfrac{DK^2}{DK\cdot DM}=\dfrac{EL^2}{EL\cdot EN} $ therefore $\dfrac{DK}{ DM}=\dfrac{EL}{ EN}.$ R HAS Your lemma is wrong if $ABC$ is isosceles. Eray wrote: In an acute triangle $ABC$, a point $P$ is taken on the $A$-altitude. Lines $BP$ and $CP$ intersect the sides $AC$ and $AB$ at points $D$ and $E$, respectively. Tangents drawn from points $D$ and $E$ to the circumcircle of triangle $BPC$ are tangent to it at points $K$ and $L$, respectively, which are in the interior of triangle $ABC$. Line $KD$ intersects the circumcircle of triangle $AKC$ at point $M$ for the second time, and line $LE$ intersects the circumcircle of triangle $ALB$ at point $N$ for the second time. Prove that\[ \frac{KD}{MD}=\frac{LE}{NE} \iff \text{Point P is the orthocenter of triangle ABC}\] So in problem we must have $ABC$ isn't isosceles you re right the triangle shouldn't be isoceles