Let $O,N,H$ be the circumcenter, nine-point center, and orthocenter of $\triangle ABC$, respectively. Let $S,M$ be the midpoints of $AB,BC$ respectively. We have $OT\perp BC$ and $TP\perp OT$, so $TP\parallel BC\parallel SM$. As $N$ is the midpoint of $OH$, and we have $HP\perp PL$ and $OT\perp PL$, We have $NP=NL$. As $S,M$ lies on the nine-point circle, we have $NS=NM$. So $N$ lies on the perpendicular bisector of both $SM$ and $TP$, and we have $SM\perp TP$. We have $SMPL$ is an isoceles trapezoid, so $S,M,P,L$ are concyclic.

I did some trig bash stuff to solve this :kekw:

Alternative solution: Let $M,N,K$ be the midpoints of $AB, BC, AC$ and $E$ the foot of the perpendicular from $B$ to $AC$. Since $MN \parallel LP \parallel EK$, the claim is equivalent to $MNLP$ being an isosceles trapezoid. But since $EKLP$ is a rectangle, this is equivalent to $MNKE$ being an isosceles trapezoid. The rest is quite trivial since $NKAM$ is a parallelogram and $EAM$ is isosceles by Thales.

Let $D,E,M$ be the midpoint of $BA,BC,AC$, respectively. Since $LA=LC$, easy angle chasing show that the tangent at $L$ of $(ABC)$ is parallel to $AC$, so it’s also parallel to $DE$. (Consequently, $DE\perp BP$.) Thus, we have to prove that $PLED$ is an isosceles trapezoid. Let $X,Y$ be the projection from $P,L$ to $DE$. It’s suffice to prove that $DX=EY$. But note that $X,Y$ are basically the projection from $A$ to $DE$ and $M$ to $DE$, respectively. We have $\triangle ADE\cup\{X\}\cong\triangle MED\cup\{Y\}$, so we’re done.

Is it doable with barycentric coordinate?

Anybody with barycentric?

This is a very easy problem but it is nice and elegant. It is probably a good tutorial for when learning angle chasing. Name the midpoints of $BC$, $AC$ and $AB$, $M$, $K$ and $N$ respectively. We also let $Q$ be the projection of $B$ onto $AC$. A very easy to prove fact is that a trapezoid is cyclic if and only if it is cyclic (this is in EGMO, the proof is quite trivial.) In this case, we clearly see that $NM \parallel AC \perp OL \perp PL$, and so $NMLP$ is a trapezoid. We can now project $P$ and $L$ onto $AC$, so that we just have to show that trapezoid $NMQK$ is isosceles which is equivalent to showing it is cyclic. This allows us to remove some of the more annoying points in our diagram. Proving the cyclicity of $NMQK$ is not too tricky. Just note that $M$ is the center of $(BQC)$, and so \[ \measuredangle MQC=\measuredangle ACB=\measuredangle MNK \]Thanks to $NKCM$ being a parallelogram due to the parallel lines.

Ijust realised that my solution is basically identical to Tintarn’s. Oh well…

Let $E$ and $D$ be the midpoints of $BC$ and $BA$. Let $O$ be the center of $(ABC)$. We get $PL||ED$ so all that is left is to show that $PLED$ is an isosceles trapezoid or $d(E,BP)=d(D,LO)$. Now consider $(BEOD)$. It is sufficient to show that $BP$ and $LO$ are equidistant from the center of $(BEOD)$ but this follows from the fact that $B$ and $O$ are antipodes and $BP||LO$.

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