gobathegreat wrote: Let $D$, $E$ and $F$ be points in which incircle of triangle $ABC$ touches sides $BC$, $CA$ and $AB$, respectively, and let $I$ be a center of that circle.Furthermore, let $P$ be a foot of perpendicular from point $I$ to line $AD$, and let $M$ be midpoint of $DE$. If $\{N\}=PM\cap{AC}$, prove that $\color{red}{ D }\normalcolor N \parallel EF$ Typo corrected My solution : Let $ T=AD \cap \odot (I) $ ( $ T \neq D $ ) . Easy to see $ A, E, F, I, P $ are concyclic . From $ \angle ENP=\angle AET=\angle EDP $ $ \Longrightarrow D, E, P, N $ are concyclic , so we get $ \angle AND=\angle EPA=\angle EFA=\angle AEF $ $ \Longrightarrow DN \parallel EF $ . Q.E.D

Dear Mathlinkers, the circle (PIMD) is also fruitfull... Sincerely Jean-Louis

My solution: $AD\cap{(I)} = T$, $PM\cap{EF} = K$ $\Rightarrow$ $DP = PT, DM = ME$ $\Rightarrow$ $PM\parallel{TE}$ (1) Now: $-1 = (EFDT) = E(EFDT) = E(NKMT)$ (2) (1), (2) $\Rightarrow$ $MK = MN$ $\Rightarrow$ $EF\parallel{DN}$ Q.E.D

Here is a solution involving no extra points to be used. Because $\angle API= \angle AEI= 90^{\circ}$ we have that $A,P,I,E$ are concyclic. From that we have $ \angle APE = \angle AIE = 90^{\circ}- \frac{\alpha}{2}$. It's easy to see that the points $I,M,C$ are collinear. From $\angle IPD = \angle IMD = 90^{\circ}$ we see that $P,I,M,D$ are concyclic. Now we have $\angle NPD = \angle MPD = \angle MID = \angle CID = 90^{\circ}-\frac{\gamma}{2}=\angle DEC= \angle DEN$, and here we see that $P,D,N,E$ are concyclic too and $\angle DNA=\angle DNE=\angle APE= 90-\frac{\alpha}{2}=\angle FEA \Rightarrow DN \parallel EF$, Q.E.D.

My solution: $AD\cap EF\equiv R$. Apply Menelaus theorem for $\triangle DAE$ has $\overline{P,M,N}\Rightarrow\frac{PD}{PA}=\frac{EN}{AN}$ We'll prove $\frac{PD}{PA}=\frac{DR}{DA}\Leftrightarrow PD^2=PA.PR\Leftrightarrow AP^2-AF^2=PA.PR\Leftrightarrow AF^2=PA.RA$ It's correct because $\triangle FAR$ and $\triangle PAF$ are similar $\Rightarrow AF^2=AR.AP$ So, $\frac{EN}{AN}=\frac{DR}{DA}\Rightarrow DN\parallel EF$

Rename $P$ as $Q$. Let $R$ be midpoint of $DE$ and $QR$ $\cap$ $AC$ $=$ $S$. We need to show, $DS||EF$. Notice, $Q$ as the midpoint of the $D-$symmedian chord. $$\angle DQS=\angle DD'E=\angle DFE=\angle DES$$Hence, $QDSE$ cyclic. Now, $$\angle AEF=\angle FDE=\angle QDE+\angle FDQ=\angle QSE + \angle QED =\angle ASD$$