Problem 2050

Lemma 1. Let $ABC$ be a triangle with incenter $I$. $IB, IC$ meet circle diameter $BC$ at $S, T$, respectively. $P$ is any point on circle diameter $BC$. $M$ is midpoint of $BC$. $MP$ meets $A$-midline at $Q$. $AQ$ meets $BC$ at $R$. $K, L$ lie on line $CS, BT$ such that $RK \perp PC, RL \perp PB$. Prove that $IP \perp KL$.

Proof of Lemma 1. Let $J$ be reflection of $R$ in $M$. Let $X, Y, Z$ be projection of $I$ on lines $BC, CP, PB$, respectively.
We see that $\triangle MEF \cup Q \sim \triangle ABC \cup J$. Let $U, V$ be incenters of triangle $JAB$ and $JAC$, respectively, then $\angle VAC = \frac{1}{2} \angle JAC = \frac{1}{2} \angle QMF = \frac{1}{2} \angle PMT = \angle ICY$, similarly $\angle UAB = \angle IBZ$, we deduce that
$$\frac{IY}{d(V, AC)} = \frac{IC}{AV} \quad \text{and} \quad \frac{IZ}{d(U, AB)} = \frac{IB}{AU}.$$
Hence,
$$\begin{aligned} \frac{IY}{IZ} &= \frac{IY}{IX} \cdot \frac{IX}{IZ} \\ &= \frac{d(V, AC) \cdot \frac{IC}{AV}}{IX} \cdot \frac{IX}{d(U, AB) \cdot \frac{IB}{AU}} \\ &= \frac{d(V, BC)}{IX} \cdot \frac{IC}{AV} \cdot \frac{IX}{d(U, AB)} \cdot \frac{AU}{IB} \\ &= \frac{CV}{CI} \cdot \frac{IC}{AV} \cdot \frac{BI}{BU} \cdot \frac{AU}{IB} \\ &= \frac{VC}{VA} \cdot \frac{UA}{UB}. \end{aligned} \quad (1)$$
Let $L'$ be reflection of $L$ in $M$. Easily seen $JL' = RL$, also $L'$ is $A$-excenter of triangle $AJC$. Thus $\triangle AJL' \sim \triangle AVC$. From this,
$$\frac{RL}{AJ} = \frac{JL'}{AJ} = \frac{VC}{VA}. \quad (2)$$
Similarly,
$$\frac{RK}{AJ} = \frac{UB}{UC}. \quad (3)$$
From (1), (2), and (3), we get
$$\frac{IY}{IZ} = \frac{RL}{RK}.$$
We obtain two similar right triangles $\triangle IPZ \sim \triangle KLR$, thus $PI \perp KL$.

Back to main problem.
Proof. Let $P'$ and $I'$ be reflections of $P$ and $I$ in $O$, respectively. We easily seen that $I'$ is incenter of triangle $P'AB$. Since $PC \parallel P'A$, the bisectors $PL$ and $AI'$ of angles $\angle CPB$ and $\angle P'AB$ are parallel. Since $BH \perp PL$, lines $BH$ and lines $AI'$ meet at $T$ on circle diameter $AB$. Similarly, lines $AK$ and lines $BI'$ meet at $S$ on circle diameter $AB$.
Now we consider triangle $P'AB$ with point $P$ on line $AB$, $I'$ is incenter of $P'AB$, $AI'$ and $BI'$ meet circle diameter $AB$ at $T$ and $S$, respectively, $O$ is a point on circle diameter $AB$ such that $O$ is midpoint of $PP'$, perpendicular lines from $P$ to $OB$, $OA$ meet $BT$, $AS$ at $H$, $K$, respectively. It follows from the particular case of lemma, we get $OI' \perp HK$. We complete the proof.

Alternative solution. Note that $PK \perp AJ = CO$, $PH \perp BL = DO$, hence the original claim that $\triangle PHK$ and $\triangle ICD$ are orthogonal is equivalent to the perpendiculars from $P, H, K$ to $CD, DI, IC$ respectively being concurrent. Let the perpendiculars of $H, K$ to $CD, DI, IC$ respectively meet at $X$, construct the point $P' \in CD$ symmetric to $P$ with respect to $O$, and let $I'$ be the incenter of $\triangle P'AB$; then we can obtain that $AK, BH$ meet at the orthocenter $H'$ of $\triangle AI'B$. Since $HH'KX$ is a parallelogram, it therefore suffices to prove that the midpoint of the projections of $H, K$ onto $AB$ equals the midpoint of the projection of $H'$ onto $AB$ and $P$, and these are respectively the tangent points $Q, R, S$ of the incircles of $\triangle PBC, \triangle PDA, \triangle P'AB$ with $BC$. Let $\overline{PA} = a$, $\overline{PB} = b$, $\overline{PC} = c$, $\overline{PD} = d$, then (directed lengths)
$$PQ = \frac{b+c-(a+b)}{2} = \frac{c-a}{2}, \quad PR = -\frac{a+d-(a+b)}{2} = \frac{b-d}{2},$$
$$PS = PB - SB = b - \frac{(a+b)+d-c}{2} = \frac{b-a+c-d}{2} = PQ + PR. \blacksquare$$