Given-a-regular-triangle-ABC-AG-2GC-CK-2KB-AK-BG-M-Prove-that-CM-AK-

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

Given a regular triangle A B C .

A G = 2 G C . C K = 2 K B . A K \cap B G = M .

Prove that C M ⊥ A K .

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{C G}{G A} \times \frac{A M}{M K} \times \frac{K B}{B C} = 1

\frac{1}{2} \times \frac{A M}{M K} \times \frac{1}{3} = 1

\Rightarrow \frac{A M}{M K} = \frac{6}{1}

\Rightarrow A M = \frac{6 \times A K}{7}

\frac{C K}{K B} \times \frac{B M}{M G} \times \frac{G A}{A C} = 1

\frac{2}{1} \times \frac{B M}{M G} \times \frac{2}{3} = 1

\Rightarrow \frac{B M}{M G} = \frac{3}{4}

\Rightarrow B M = \frac{3 \times B G}{7}

s i n c e B G = A K,

\Rightarrow B M = \frac{A M}{2}

α = α_{1} + α_{2}

β = α_{1} + β_{1}

m a k e Δ A B N \equiv Δ B C M

∠ M B N = α_{1} + α_{2} = 60 °

Δ B M N i s e q u i l a t e r a l .

B N / / M A

∠ A M N = 60 °

M N = B M = \frac{A M}{2}

\Rightarrow ∠ M A N = α_{1} + β_{1} = 30 °

∠ A M C = α + β = α_{1} + α_{2} + α_{1} + β_{1}

= 60 ° + α_{1} + β_{1}

= 60 ° + 30 ° = 90 °

\Rightarrow C M ⊥ A K

Answered by A5T last updated on 23/Mar/25

Commented by A5T last updated on 23/Mar/25

WLOG, let AB = BC = CA = 3 \Rightarrow BK = 1

\frac{AD}{DB} \times \frac{BK}{KC} \times \frac{CG}{GA} = 1 \Rightarrow \frac{AD}{DB} = 4 \Rightarrow AD = \frac{12}{5}

AK = \sqrt{{BK}^{2} + {AB}^{2} - 2 BK \times ABcos 60} = \sqrt{7}

\frac{AM}{MK} = \frac{AD}{DB} + \frac{AG}{GC} = 4 + 2 = 6 \Rightarrow AM = \frac{6 \sqrt{7}}{7}

CD = \sqrt{{AD}^{2} + {AC}^{2} - 2 AD \times ACcos 60} = \sqrt{\frac{189}{25}}

\frac{CM}{MD} = \frac{CG}{GA} + \frac{CK}{CB} = \frac{5}{2} \Rightarrow DM = \frac{2 \sqrt{189}}{7 \times 5}

{AM}^{2} + {DM}^{2} = \frac{4 \times 27}{7 \times 5^{2}} + \frac{36}{7} = \frac{144}{25} = {AD}^{2}

\Rightarrow ∠ AMD = 90 ° \Rightarrow ∠ AMC = 90 ° \Rightarrow CM ⊥ AK

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