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Since ABA₁B₁ is concyclic, with AB as diameter, AA₁ and BB₁ are two chords intersecting at H, thus AH×HA₁ = BH×HB₁. Similarly consider cyclic quadrilateral BCB₁C₁, we have BH×HB₁ = CH×HC₁. Hence, AH×HA₁ = BH×HB₁ = CH×HC₁.
triunghiul ANB1este asemenea cu triunghiul BHA1,BHC1~CHB1 (UU)=>AH/BH=HB1/HA1;HB/HC=HC1/HB1 => AH×HA₁ = BH×HB₁ = CH×HC₁.
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Since ABA₁B₁ is concyclic, with AB as diameter,
ReplyDeleteAA₁ and BB₁ are two chords intersecting at H,
thus AH×HA₁ = BH×HB₁.
Similarly consider cyclic quadrilateral BCB₁C₁,
we have BH×HB₁ = CH×HC₁.
Hence, AH×HA₁ = BH×HB₁ = CH×HC₁.
triunghiul ANB1este asemenea cu triunghiul BHA1,BHC1~CHB1 (UU)=>AH/BH=HB1/HA1;HB/HC=HC1/HB1 => AH×HA₁ = BH×HB₁ = CH×HC₁.
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