Question Number 217931 by CrispyXYZ last updated on 23/Mar/25
$$\mathrm{Given}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{triangle}\:{ABC}. \\ $$$${AG}=\mathrm{2}{GC}.\:{CK}=\mathrm{2}{KB}.\:{AK}\cap{BG}={M}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{CM}\bot{AK}. \\ $$
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
$$\frac{{CG}}{{GA}}×\frac{{AM}}{{MK}}×\frac{{KB}}{{BC}}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{{AM}}{{MK}}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{AM}}{{MK}}=\frac{\mathrm{6}}{\mathrm{1}} \\ $$$$\Rightarrow{AM}=\frac{\mathrm{6}×{AK}}{\mathrm{7}} \\ $$$$\frac{{CK}}{{KB}}×\frac{{BM}}{{MG}}×\frac{{GA}}{{AC}}=\mathrm{1} \\ $$$$\frac{\mathrm{2}}{\mathrm{1}}×\frac{{BM}}{{MG}}×\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{BM}}{{MG}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{BM}=\frac{\mathrm{3}×{BG}}{\mathrm{7}} \\ $$$${since}\:{BG}={AK}, \\ $$$$\Rightarrow{BM}=\frac{{AM}}{\mathrm{2}} \\ $$$$ \\ $$$$\alpha=\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} \\ $$$$\beta=\alpha_{\mathrm{1}} +\beta_{\mathrm{1}} \\ $$$${make}\:\Delta{ABN}\equiv\Delta{BCM} \\ $$$$\angle{MBN}=\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} =\mathrm{60}° \\ $$$$\Delta{BMN}\:{is}\:{equilateral}. \\ $$$${BN}//{MA} \\ $$$$\angle{AMN}=\mathrm{60}° \\ $$$${MN}={BM}=\frac{{AM}}{\mathrm{2}} \\ $$$$\Rightarrow\angle{MAN}=\alpha_{\mathrm{1}} +\beta_{\mathrm{1}} =\mathrm{30}° \\ $$$$\angle{AMC}=\alpha+\beta=\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} +\alpha_{\mathrm{1}} +\beta_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{60}°+\alpha_{\mathrm{1}} +\beta_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{60}°+\mathrm{30}°=\mathrm{90}° \\ $$$$\Rightarrow{CM}\bot{AK} \\ $$
Answered by A5T last updated on 23/Mar/25
Commented by A5T last updated on 23/Mar/25
$$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=\mathrm{3}\Rightarrow\mathrm{BK}=\mathrm{1} \\ $$$$\frac{\mathrm{AD}}{\mathrm{DB}}×\frac{\mathrm{BK}}{\mathrm{KC}}×\frac{\mathrm{CG}}{\mathrm{GA}}=\mathrm{1}\Rightarrow\frac{\mathrm{AD}}{\mathrm{DB}}=\mathrm{4}\Rightarrow\mathrm{AD}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\mathrm{AK}=\sqrt{\mathrm{BK}^{\mathrm{2}} +\mathrm{AB}^{\mathrm{2}} −\mathrm{2BK}×\mathrm{ABcos60}}=\sqrt{\mathrm{7}} \\ $$$$\frac{\mathrm{AM}}{\mathrm{MK}}=\frac{\mathrm{AD}}{\mathrm{DB}}+\frac{\mathrm{AG}}{\mathrm{GC}}=\mathrm{4}+\mathrm{2}=\mathrm{6}\Rightarrow\mathrm{AM}=\frac{\mathrm{6}\sqrt{\mathrm{7}}}{\:\mathrm{7}} \\ $$$$\mathrm{CD}=\sqrt{\mathrm{AD}^{\mathrm{2}} +\mathrm{AC}^{\mathrm{2}} −\mathrm{2AD}×\mathrm{ACcos60}}=\sqrt{\frac{\mathrm{189}}{\mathrm{25}}} \\ $$$$\frac{\mathrm{CM}}{\mathrm{MD}}=\frac{\mathrm{CG}}{\mathrm{GA}}+\frac{\mathrm{CK}}{\mathrm{CB}}=\frac{\mathrm{5}}{\mathrm{2}}\Rightarrow\mathrm{DM}=\frac{\mathrm{2}\sqrt{\mathrm{189}}}{\:\mathrm{7}×\mathrm{5}} \\ $$$$\mathrm{AM}^{\mathrm{2}} +\mathrm{DM}^{\mathrm{2}} =\frac{\mathrm{4}×\mathrm{27}}{\mathrm{7}×\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{36}}{\mathrm{7}}=\frac{\mathrm{144}}{\mathrm{25}}=\mathrm{AD}^{\mathrm{2}} \\ $$$$\Rightarrow\angle\mathrm{AMD}=\mathrm{90}°\Rightarrow\angle\mathrm{AMC}=\mathrm{90}°\Rightarrow\mathrm{CM}\bot\mathrm{AK} \\ $$