Let $\mathcal{H}$ be the circumrectangular hyperbola passing through $P$. Suppose $H_A$ is the orthocenter of $\triangle PBC$, and define $H_B,H_C$ analogously. Then $H,H_A,H_B,H_C \in \mathcal{H}$. Also, one can easily see that $APH_AH$, et al. are parallelograms. In particular, $HH_A \parallel AP$, and so $D \in HH_A$ (with similar results for $E,F$). Then Pascal on $HH_AABCH_C$ gives that line $DF$ passes through $AH_A \cap CH_C$. But, due to the parallelogram condition, $M$ is the midpoint of segments $AH_A,BH_B$ and $CH_C$ also. Thus, $DF$ passes through $M$. Similarly, $M \in DE$, and so $D,E,F,M$ are collinear.

Moving point kills the problem.

Can be bashed. We get $D=\frac{bc(h+p)}{bc-ap}$ and others analogously. So we need to show collinearity of $D=\frac{ba(h+p)}{ba-cp} , \frac{ac(h+p)}{ac-bp}$ and $ \frac{h+p}{2}$. Divide with $h+p$ and then its trivial.

Sugiyem wrote: Moving point kills the problem. Solution please!

Here is a solution which gives the direction of this line. Let the circumcenter of $\triangle ABC$ be $O$. We only need to prove that $\angle OMD=90^\circ$. Denote the midpoint of $BC , AP$ and the antipode of $A$ by $L , N$ and $A'$ respectively. Then it's obvious that $LM \bot AP$, $ONLM$ is a parallelogram. Let the projection of $O , D , A$ to $LM$ be $U , V , W$ respectively. By $AP//HD$ and $M$ is the midpoint of $HP$, we can see $WM=VW$. However, $UW=ON=LM=LW+MV$, it means that $UL=MV$, and now it's trivial to see that $L , M$ are both lying on the circle with a diameter $OD$. So $\angle OMD=90^\circ$. Q.E.D.

Let $ O $ be the circumcenter of $ \triangle ABC, $ $ V $ be the second intersection of $ AH $ with $ \odot (O) $ and $ J, K $ be the midpoint of $ HV, PV, $ respectively, then $ JV \stackrel{\parallel}{=} MK $ and $ \triangle OVK \stackrel{-}{\sim} \triangle HDJ \stackrel{-}{\sim} \triangle VDJ \Longrightarrow \triangle OVK \stackrel{+}{\sim} \triangle VDJ, $ so $$ \measuredangle (OV, VD) = \measuredangle (OK,VJ) = \measuredangle (OK,KM) \qquad \text{and}\qquad \frac{OV}{VD} = \frac{OK}{VJ} = \frac{OK}{KM} $$$ \Longrightarrow $ $ \triangle OVD \stackrel{+}{\sim} \triangle OKM, $ hence we conclude that $ \triangle OMD \stackrel{+}{\sim} \triangle OKV $ and $ OM \perp MD. $ $ \qquad \blacksquare $

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WLOG we set $(ABC)$ as a unit circle in the complex plane centered at $O=0$. $D$ lies on $BC$ iff $d+\overline{d}bc=b+c$, hence $\overline{d}=\frac{b+c-d}{bc}$. $AP\parallel HD$ iff $\overline{\big( \frac{d-h}{p-a} \big)}=\frac{d-h}{p-a}$. Substituting $\overline{d}$ from the previous equation, we obtain $d=(a+b+c+p)\frac{bc}{bc-ap}$. Similarly $e=(a+b+c+p)\frac{ca}{ca-bp}$ and $f=(a+b+c+p)\frac{ab}{ab-cp}$. Also, $\overline{d}=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{p})\frac{ap}{ap-bc}$. $\overline{e}$ and $\overline{f}$ are cyclical. Now we evaluate the determinant $\begin{vmatrix} d && \overline{d} && 1 \\ e && \overline{e} && 1 \\ f && \overline{f} && 1 \\ \end{vmatrix} = (a+b+c+p)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{p}) \begin{vmatrix} \frac{bc}{bc-ap} && \frac{ap}{ap-bc} && 1\\ \frac{ca}{ca-bp} && \frac{bp}{bp-ca} && 1\\ \frac{ab}{ab-cp} && \frac{cp}{cp-ab} && 1\\ \end{vmatrix} =0 $ Also, $ \begin{vmatrix} d && \overline{d} &&1\\ e && \overline{e} &&1\\ m && \overline{m} &&1\\ \end{vmatrix} = (a+b+c+p)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{p}) \begin{vmatrix} \frac{bc}{bc-ap} && \frac{ap}{ap-bc} && 1\\ \frac{ca}{ca-bp} && \frac{bp}{bp-ca} && 1\\ \frac{1}{2} && \frac{1}{2} && 1\\ \end{vmatrix} =0 $ Hence $D$, $E$, $F$ are collinear and $D$, $E$, and $M$ are collinear, which suffices to prove the statement of the problem. $\square$

dagezjm wrote: Here is a solution which gives the direction of this line. Let the circumcenter of $\triangle ABC$ be $O$. We only need to prove that $\angle OMD=90^\circ$. Denote the midpoint of $BC , AP$ and the antipode of $A$ by $L , N$ and $A'$ respectively. Then it's obvious that $LM \bot AP$, $ONLM$ is a parallelogram. Let the projection of $O , D , A$ to $LM$ be $U , V , W$ respectively. By $AP//HD$ and $M$ is the midpoint of $HP$, we can see $WM=VW$. However, $UW=ON=LM=LW+MV$, it means that $UL=MV$, and now it's trivial to see that $L , M$ are both lying on the circle with a diameter $OD$. So $\angle OMD=90^\circ$. Q.E.D. [asy][asy] size(150); pair A=dir(130),B=dir(-160),C=dir(-20),H=A+B+C,P=dir(-93),M=1/2*(H+P),L=1/2*(B+C); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(-120)); dot("$L$",L,dir(L)); dot("$M$",M,dir(M)); dot("$P$",P,dir(P)); dot(extension(A,P,L,M)); draw(unitcircle,red); draw(A--B--C--A,blue); draw(H--P,brown); draw(L--M^^A--P,green); markscalefactor=0.01; draw(rightanglemark(P,extension(A,P,L,M),L)); markscalefactor=0.008; add(pathticks(P--M,1)); add(pathticks(M--H,1)); [/asy][/asy] Can someone please tell why is $LM \perp AP$. Like, this can of course be proven using complex numbers, but how to prove it synthetically ?

guptaamitu1 wrote: dagezjm wrote: Here is a solution which gives the direction of this line. Let the circumcenter of $\triangle ABC$ be $O$. We only need to prove that $\angle OMD=90^\circ$. Denote the midpoint of $BC , AP$ and the antipode of $A$ by $L , N$ and $A'$ respectively. Then it's obvious that $LM \bot AP$, $ONLM$ is a parallelogram. Let the projection of $O , D , A$ to $LM$ be $U , V , W$ respectively. By $AP//HD$ and $M$ is the midpoint of $HP$, we can see $WM=VW$. However, $UW=ON=LM=LW+MV$, it means that $UL=MV$, and now it's trivial to see that $L , M$ are both lying on the circle with a diameter $OD$. So $\angle OMD=90^\circ$. Q.E.D. [asy][asy] size(150); pair A=dir(130),B=dir(-160),C=dir(-20),H=A+B+C,P=dir(-93),M=1/2*(H+P),L=1/2*(B+C); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(-120)); dot("$L$",L,dir(L)); dot("$M$",M,dir(M)); dot("$P$",P,dir(P)); dot(extension(A,P,L,M)); draw(unitcircle,red); draw(A--B--C--A,blue); draw(H--P,brown); draw(L--M^^A--P,green); markscalefactor=0.01; draw(rightanglemark(P,extension(A,P,L,M),L)); markscalefactor=0.008; add(pathticks(P--M,1)); add(pathticks(M--H,1)); [/asy][/asy] Can someone please tell why is $LM \perp AP$. Like, this can of course be proven using complex numbers, but how to prove it synthetically ? You use the homothety at $H$ with ratio $\frac{1}{2}$ and use euler circle