Let $ABC$ be an acute triangle with an altitude $AD$. Let $H$ be the orthocenter of triangle $ABC$. The circle $\Omega$ passes through the points $A$ and $B$, and touches the line $AC$. Let $BE$ be the diameter of $\Omega$. The lines~$BH$ and $AH$ intersect $\Omega$ for the second time at points $K$ and $L$, respectively. The lines $EK$ and $AB$ intersect at the point~$T$. Prove that $\angle BDK=\angle BLT$.
Problem
Source: Kazakhstan National Olympiad 2024 (10-11 grade), P1
Tags: geometry
bin_sherlo
21.03.2024 09:29
$AD\cap EK=S$ $\angle BDS=90=\angle BKS\implies B,D,K,S$ are cyclic. We have $EK\perp BH\perp AC\implies EK\parallel AC$ $\angle ABK=\angle CAK=\angle EKA$ and $\angle KAS=\angle KAL=\angle KBL$ Hence $180-\angle LST=\angle KSL=\angle KAS+\angle SKA=\angle KBL+\angle ABK=\angle ABL$ $\implies T,S,L,B$ are cyclic. $\angle BDK=\angle BST=\angle BLT$ a desired.$\blacksquare$
sami1618
08.06.2024 21:52
Claim: Triangles $BDL$ and $BKT$ are similar $$\angle BDL=\angle BKT=90^{\circ}$$$$\angle LBD=\angle ALB-90^{\circ}=90^{\circ}-\angle AEB=90^{\circ}-\angle BAC=\angle ABH=\angle TBK$$ Then by SST triangles $BDK$ and $BLE$ are also similar, proving the equality.
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Captainscrubz
26.08.2024 15:21
Prove ET parallel to AC and then prove triangle TBL is similar to triangle KBD
mahmudlusenan
05.11.2024 20:41
it is easy to see that $\angle DBH = \angle CAH = \angle LBT$ so $\angle DBL = \angle KBT $ and $\angle BDL = \angle BKT$ then $\triangle BDL\sim\triangle BKT$ then by the spiral similarity lemma $\triangle BDK\sim\triangle BLT$ so $\angle BDK = \angle BLT$