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Question Number 217931 by CrispyXYZ last updated on 23/Mar/25

Given a regular triangle ABC. AG=2GC. CK=2KB. AK∩BG=M. Prove that CM⊥AK.

$$\mathrm{Given}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{triangle}\:{ABC}. \\ $$$${AG}=\mathrm{2}{GC}.\:{CK}=\mathrm{2}{KB}.\:{AK}\cap{BG}={M}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{CM}\bot{AK}. \\ $$

Commented by CrispyXYZ last updated on 23/Mar/25

Answered by mr W last updated on 23/Mar/25

Commented by mr W last updated on 23/Mar/25

((CG)/(GA))×((AM)/(MK))×((KB)/(BC))=1 (1/2)×((AM)/(MK))×(1/3)=1 ⇒((AM)/(MK))=(6/1) ⇒AM=((6×AK)/7) ((CK)/(KB))×((BM)/(MG))×((GA)/(AC))=1 (2/1)×((BM)/(MG))×(2/3)=1 ⇒((BM)/(MG))=(3/4) ⇒BM=((3×BG)/7) since BG=AK, ⇒BM=((AM)/2) α=α_1 +α_2 β=α_1 +β_1 make ΔABN≡ΔBCM ∠MBN=α_1 +α_2 =60° ΔBMN is equilateral. BN//MA ∠AMN=60° MN=BM=((AM)/2) ⇒∠MAN=α_1 +β_1 =30° ∠AMC=α+β=α_1 +α_2 +α_1 +β_1 =60°+α_1 +β_1 =60°+30°=90° ⇒CM⊥AK

$$\frac{{CG}}{{GA}}×\frac{{AM}}{{MK}}×\frac{{KB}}{{BC}}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{{AM}}{{MK}}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{AM}}{{MK}}=\frac{\mathrm{6}}{\mathrm{1}} \\ $$$$\Rightarrow{AM}=\frac{\mathrm{6}×{AK}}{\mathrm{7}} \\ $$$$\frac{{CK}}{{KB}}×\frac{{BM}}{{MG}}×\frac{{GA}}{{AC}}=\mathrm{1} \\ $$$$\frac{\mathrm{2}}{\mathrm{1}}×\frac{{BM}}{{MG}}×\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{BM}}{{MG}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{BM}=\frac{\mathrm{3}×{BG}}{\mathrm{7}} \\ $$$${since}\:{BG}={AK}, \\ $$$$\Rightarrow{BM}=\frac{{AM}}{\mathrm{2}} \\ $$$$ \\ $$$$\alpha=\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} \\ $$$$\beta=\alpha_{\mathrm{1}} +\beta_{\mathrm{1}} \\ $$$${make}\:\Delta{ABN}\equiv\Delta{BCM} \\ $$$$\angle{MBN}=\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} =\mathrm{60}° \\ $$$$\Delta{BMN}\:{is}\:{equilateral}. \\ $$$${BN}//{MA} \\ $$$$\angle{AMN}=\mathrm{60}° \\ $$$${MN}={BM}=\frac{{AM}}{\mathrm{2}} \\ $$$$\Rightarrow\angle{MAN}=\alpha_{\mathrm{1}} +\beta_{\mathrm{1}} =\mathrm{30}° \\ $$$$\angle{AMC}=\alpha+\beta=\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} +\alpha_{\mathrm{1}} +\beta_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{60}°+\alpha_{\mathrm{1}} +\beta_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{60}°+\mathrm{30}°=\mathrm{90}° \\ $$$$\Rightarrow{CM}\bot{AK} \\ $$

Answered by A5T last updated on 23/Mar/25

Commented by A5T last updated on 23/Mar/25

WLOG, let AB=BC=CA=3⇒BK=1 ((AD)/(DB))×((BK)/(KC))×((CG)/(GA))=1⇒((AD)/(DB))=4⇒AD=((12)/5) AK=(√(BK^2 +AB^2 −2BK×ABcos60))=(√7) ((AM)/(MK))=((AD)/(DB))+((AG)/(GC))=4+2=6⇒AM=((6(√7))/( 7)) CD=(√(AD^2 +AC^2 −2AD×ACcos60))=(√((189)/(25))) ((CM)/(MD))=((CG)/(GA))+((CK)/(CB))=(5/2)⇒DM=((2(√(189)))/( 7×5)) AM^2 +DM^2 =((4×27)/(7×5^2 ))+((36)/7)=((144)/(25))=AD^2 ⇒∠AMD=90°⇒∠AMC=90°⇒CM⊥AK

$$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=\mathrm{3}\Rightarrow\mathrm{BK}=\mathrm{1} \\ $$$$\frac{\mathrm{AD}}{\mathrm{DB}}×\frac{\mathrm{BK}}{\mathrm{KC}}×\frac{\mathrm{CG}}{\mathrm{GA}}=\mathrm{1}\Rightarrow\frac{\mathrm{AD}}{\mathrm{DB}}=\mathrm{4}\Rightarrow\mathrm{AD}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\mathrm{AK}=\sqrt{\mathrm{BK}^{\mathrm{2}} +\mathrm{AB}^{\mathrm{2}} −\mathrm{2BK}×\mathrm{ABcos60}}=\sqrt{\mathrm{7}} \\ $$$$\frac{\mathrm{AM}}{\mathrm{MK}}=\frac{\mathrm{AD}}{\mathrm{DB}}+\frac{\mathrm{AG}}{\mathrm{GC}}=\mathrm{4}+\mathrm{2}=\mathrm{6}\Rightarrow\mathrm{AM}=\frac{\mathrm{6}\sqrt{\mathrm{7}}}{\:\mathrm{7}} \\ $$$$\mathrm{CD}=\sqrt{\mathrm{AD}^{\mathrm{2}} +\mathrm{AC}^{\mathrm{2}} −\mathrm{2AD}×\mathrm{ACcos60}}=\sqrt{\frac{\mathrm{189}}{\mathrm{25}}} \\ $$$$\frac{\mathrm{CM}}{\mathrm{MD}}=\frac{\mathrm{CG}}{\mathrm{GA}}+\frac{\mathrm{CK}}{\mathrm{CB}}=\frac{\mathrm{5}}{\mathrm{2}}\Rightarrow\mathrm{DM}=\frac{\mathrm{2}\sqrt{\mathrm{189}}}{\:\mathrm{7}×\mathrm{5}} \\ $$$$\mathrm{AM}^{\mathrm{2}} +\mathrm{DM}^{\mathrm{2}} =\frac{\mathrm{4}×\mathrm{27}}{\mathrm{7}×\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{36}}{\mathrm{7}}=\frac{\mathrm{144}}{\mathrm{25}}=\mathrm{AD}^{\mathrm{2}} \\ $$$$\Rightarrow\angle\mathrm{AMD}=\mathrm{90}°\Rightarrow\angle\mathrm{AMC}=\mathrm{90}°\Rightarrow\mathrm{CM}\bot\mathrm{AK} \\ $$

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