$ABC$ triangle with $|AB|<|BC|<|CA|$ has the incenter $I$. The orthocenters of triangles $IBC, IAC$ and $IAB$ are $H_A, H_A$ and $H_A$. $H_BH_C$ intersect $BC$ at $K_A$ and perpendicular line from $I$ to $H_BH_B$ intersect $BC$ at $L_A$. $K_B, L_B, K_C, L_C$ are defined similarly. Prove that $$|K_AL_A|=|K_BL_B|+|K_CL_C|$$
Problem
Source: Turkey TST 2022 P8 Day 3
Tags: geometry, geometry proposed, orthocenter, incircle
13.03.2022 14:54
How many problems are in a Turkey TST?
13.03.2022 14:55
There are 3 days each containing exactly 3 problems which gives a total of 9 problems.
13.03.2022 14:56
Attachments:

13.03.2022 16:53
Lemma 1 $H_{B},D,H_{C}$ are collinear. Proof 1 Since $BH_{C}$ and $CH_{B}$ are parallel and $\frac{BD}{DC}=\frac{BF}{CE}=\frac{BH_{C}}{CH_{B}}$ $H_{B},D,H_{C}$ are collinear as desired. Lemma 2 Let M be the midpoint of $BC$. Then $IM\perp H_{B}H_{C}$ Proof 2 Let $AI\cap BH_{C}=X$ and $AI\cap CH_{B}=Y$. It is easy to see that $IXDB$ are cyclic. Let $AI\cap (ABC)=N$. It is known that $NI=NB=NC$ and $BXMN$ are cyclic. Then $\angle{DBX}=\angle{MNI}$. Furthermore since the triangles $BFH_{C}$ and $NMC$, $\frac{BD}{BH_{C}}=\frac{BF}{BH_{C}}=\frac{NM}{NC}=\frac{NM}{NI}$ Thus $MIN \sim DH_{C}B \implies \angle{NIM}=\angle{BH_{C}D}$. Let $IM\cap H_{B}H_{C}=P$. Then $IXPH_{C}$ is cyclic $\implies$ $IP \perp H_{B}H_{C}$. From this, we can immediately see $K_{A}=D$ and $L_{A}=M_{A}$ where $M_{A}$ is the midpoint of $BC$. Similar for other points and the conclusion is obvious.
13.03.2022 18:01
Here is an amazing approach to this problem by serdarbozdag, hakN, sevket12, cookierockie, bariskoyuncu, infinityfun: Let $M_A,M_B,M_C$ be the midpoints of $BC,AC,AB$ respectively and incircle of $\triangle ABC$ touches $BC,AC,AB$ at $D,E,F$ respectively. Claim: $H_BH_C$ is the polar of $M_A$ wrt incircle. Proof: Let $EF\cap M_AM_C=N.$ Then by iran lemma we know $C,I,N$ are collinear and $IC\perp AH_B.$ Also we know that $N$ lies on the polar of $A$ wrt incircle. Thus, $AH_B$ is the polar of $N$ wrt incircle. By La-Hire's Theorem, $N$ lies on the polar of $H_B.$ Which implies $M_CM_A$ is the polar of $H_B$ because $H_BI\perp M_CM_A.$ Similarly we can get $M_BM_A$ is the polar of $H_C.$ So again by La-Hire's Theorem, $H_BH_C$ is the polar of $M_A$ wrt incircle. After this claim we can easily say $L_A=M_A$ and $K_A=D$ (because $M_AD$ is tangent to incircle.) And the rest follows.
13.03.2022 18:18
For generalization and more properties see https://artofproblemsolving.com/community/c6h1252050p6455626.