Let ABC be an acute triangle with an altitude AD. Let H be the orthocenter of triangle ABC. The circle Ω passes through the points A and B, and touches the line AC. Let BE be the diameter of Ω. The lines~BH and AH intersect Ω for the second time at points K and L, respectively. The lines EK and AB intersect at the point~T. Prove that ∠BDK=∠BLT.
Problem
Source: Kazakhstan National Olympiad 2024 (10-11 grade), P1
Tags: geometry
bin_sherlo
21.03.2024 09:29
AD∩EK=S ∠BDS=90=∠BKS⟹B,D,K,S are cyclic. We have EK⊥BH⊥AC⟹EK∥AC ∠ABK=∠CAK=∠EKA and ∠KAS=∠KAL=∠KBL Hence 180−∠LST=∠KSL=∠KAS+∠SKA=∠KBL+∠ABK=∠ABL ⟹T,S,L,B are cyclic. ∠BDK=∠BST=∠BLT a desired.◼
sami1618
08.06.2024 21:52
Claim: Triangles BDL and BKT are similar ∠BDL=∠BKT=90∘∠LBD=∠ALB−90∘=90∘−∠AEB=90∘−∠BAC=∠ABH=∠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 ∠DBH=∠CAH=∠LBT so ∠DBL=∠KBT and ∠BDL=∠BKT then △BDL∼△BKT then by the spiral similarity lemma △BDK∼△BLT so ∠BDK=∠BLT