Nice problem General problem Given triangle $\triangle ABC$ and let $D$ be the foot of angle bisectors of $A$. If $X \in AD, BX \cap AC \equiv E, CX \cap AB \equiv F$ $M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$. Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$. Show that: $DK=DL$ Proof $Y \equiv EF \cap BC, R \equiv AH \cap EF$ let $W,Z$ the proyections of $A,D$ at $EF$ respectively. Denote $P$ a point at $ZD$ such that $\odot (APD)$ is tangent to $AZ$, and $Q \equiv EF \cap AP$ Is clear $\angle AYF =\angle ADP =\angle PAZ \Rightarrow AZ^2=ZY.ZQ=ZP.ZD \Rightarrow YQPD$ is cyclical $\Rightarrow \angle ZYD=\angle APZ \Rightarrow \triangle AZP \cup Q \sim \triangle HWA \cup R \Rightarrow \frac{AR}{RH}=\frac{PQ}{QA}$ and $\angle ARN=\angle MQA$ Denote $K'$ and $L'$ points at $EF$ such that $AN \parallel PK'$ and $AM \parallel PL' \Rightarrow AMN \cup H \sim PK'L' \cup A \Rightarrow \triangle HMN \sim \triangle AK'L' \Rightarrow K' \equiv K$ and $L \equiv L'$ is evident that $KP=PL \Rightarrow DK=DL$

First we see that K,L can be construct uniquely by let the symmetric of H wrt EF be H' and draw two parallel line from A to H'M, H'N we have KL. The same idea as IMO 2018 G6, we will reconstruct it. Let the altitude of A,D and BC meet EF at U,V,W and P lies on DU such that UPA~UAD then UP.UD=UA^2 and UDA=UWA=UAP so UW.UQ=UD.UP gives us WQPD cyclic. Cause AVH=AUP=180-AWD and APU=UAD=VWH=VAR so AUP ~ HVA and PUQ=RVA=90 then AR/RH=PQ/QA + ARN=KQA. Re-construct K,L as PK//AN, PL//AM so AMN-H~PKL-A so AKL~HMN. (Q.E.D)

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If we get $ T , P , Y $ foot of $ A , D , Y $ with $ EF$ and $Q , X$ intersection of $AH , AD$ with $EF$ . Get $N$ the point on $TE$ and $L' , K'$ reflection of $L , K$ about $T$ . $HN||AL' , HM||AK' \rightarrow \frac{NQ}{NT+TL}=\frac{NQ}{ML}=\frac{HQ}{HA}=\frac{MQ}{NK}=\frac{2NT-NQ}{NT+TK}$ $\rightarrow NT+TL=\frac{AH}{HQ}.NQ , NT+TK=\frac{AH}{HQ}.(2NT-NQ) \rightarrow |TL-TK|=2.\frac{AH}{HQ}.|NQ-NT| = 2.\frac{AH}{HQ}.QT$ so the problem is equal $\leftrightarrow PL=PK \leftrightarrow |TL-TK|=2TP \leftrightarrow TP=\frac{AH}{HQ}.QT $ If $D'$ intersection of $EF , BC$ we can see that $\angle{DAD'}=90$ so $ATHD' , PDD'A$ are cyclic . $\frac{AH}{HQ}.QT=AH.\frac{QT}{QA}.\frac{AQ}{QH}=AH . \sin{BD'F} . \frac{\sin{AD'F}}{\sin{BD'F}}.\frac{AD'}{D'H} = AH . \sin{AD'F} . \frac{AD}{AH} = \sin{AD'F} . AD = TP $ and we're done $\blacksquare$