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

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

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

$$\frac{CG}{GA}\times \frac{AM}{MK}\times \frac{KB}{BC}=1$$$$\frac{1}{2}\times \frac{AM}{MK}\times \frac{1}{3}=1$$$$\Rightarrow \frac{AM}{MK}=\frac{6}{1}$$$$\Rightarrow AM=\frac{6\times AK}{7}$$$$\frac{CK}{KB}\times \frac{BM}{MG}\times \frac{GA}{AC}=1$$$$\frac{2}{1}\times \frac{BM}{MG}\times \frac{2}{3}=1$$$$\Rightarrow \frac{BM}{MG}=\frac{3}{4}$$$$\Rightarrow BM=\frac{3\times BG}{7}$$$$sinceBG=AK,$$$$\Rightarrow BM=\frac{AM}{2}$$$$$$$$\alpha ={\alpha }_1+{\alpha }_2$$$$\beta ={\alpha }_1+{\beta }_1$$$$make\mathrm{\Delta }ABN\equiv \mathrm{\Delta }BCM$$$$\mathrm{\angle }MBN={\alpha }_1+{\alpha }_2=60°$$$$\mathrm{\Delta }BMNisequilateral.$$$$BN//MA$$$$\mathrm{\angle }AMN=60°$$$$MN=BM=\frac{AM}{2}$$$$\Rightarrow \mathrm{\angle }MAN={\alpha }_1+{\beta }_1=30°$$$$\mathrm{\angle }AMC=\alpha +\beta ={\alpha }_1+{\alpha }_2+{\alpha }_1+{\beta }_1$$$$=60°+{\alpha }_1+{\beta }_1$$$$=60°+30°=90°$$$$\Rightarrow CM\mathrm{\perp }AK$$

Answered by A5T last updated on 23/Mar/25

Commented by A5T last updated on 23/Mar/25

$$\mathrm{WLOG},\mathrm{let}\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=3\Rightarrow \mathrm{BK}=1$$$$\frac{\mathrm{AD}}{\mathrm{DB}}\times \frac{\mathrm{BK}}{\mathrm{KC}}\times \frac{\mathrm{CG}}{\mathrm{GA}}=1\Rightarrow \frac{\mathrm{AD}}{\mathrm{DB}}=4\Rightarrow \mathrm{AD}=\frac{12}{5}$$$$\mathrm{AK}=\sqrt{{\mathrm{BK}}^2+{\mathrm{AB}}^2-2\mathrm{BK}\times \mathrm{ABcos}60}=\sqrt{7}$$$$\frac{\mathrm{AM}}{\mathrm{MK}}=\frac{\mathrm{AD}}{\mathrm{DB}}+\frac{\mathrm{AG}}{\mathrm{GC}}=4+2=6\Rightarrow \mathrm{AM}=\frac{6\sqrt{7}}{7}$$$$\mathrm{CD}=\sqrt{{\mathrm{AD}}^2+{\mathrm{AC}}^2-2\mathrm{AD}\times \mathrm{ACcos}60}=\sqrt{\frac{189}{25}}$$$$\frac{\mathrm{CM}}{\mathrm{MD}}=\frac{\mathrm{CG}}{\mathrm{GA}}+\frac{\mathrm{CK}}{\mathrm{CB}}=\frac{5}{2}\Rightarrow \mathrm{DM}=\frac{2\sqrt{189}}{7\times 5}$$$${\mathrm{AM}}^2+{\mathrm{DM}}^2=\frac{4\times 27}{7\times 5^2}+\frac{36}{7}=\frac{144}{25}={\mathrm{AD}}^2$$$$\Rightarrow \mathrm{\angle }\mathrm{AMD}=90°\Rightarrow \mathrm{\angle }\mathrm{AMC}=90°\Rightarrow \mathrm{CM}\mathrm{\perp }\mathrm{AK}$$

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