01/07/2026

$\triangle{ABC}, \triangle{ADE}, \triangle{AFG}$ are equilateral triangles. $K, M, N$ are the midpoint of $BD, EF, CG$ respectively. Show that $\triangle{KMN}$ is equilateral.

Prove:

\(\begin{multline} \shoveleft \text{Extend }AK, AM, AN \text{ to }K', M', N'\\ \shoveleft \text{ such that }KK'=AK, MM'=AM, NN'=AN\\ \shoveleft \implies AGN'C, ADK'B, AEM'F \text{ are all parallelograms}\\ \shoveleft \text{Then make }ABE'E \text{ to be parallelogram}\\ \shoveleft \implies ABK'D, DK'E'E \text{ are also parallelograms}\\ \shoveleft \implies \triangle{BK'E'} \text{ is also equilateral triangle}\\ \shoveleft \implies BK'=E'K', E'E=AB=BC, \angle{KBC}=240^{\circ}-\angle{ABE'}\\ \shoveleft =60^{\circ}+(180^{\circ}-\angle{ABE'})=60^{\circ}+\angle{EE'K'}=\angle{EE'K'}\\ \shoveleft \implies \triangle{EE'K'}\cong\triangle{CBK'}\implies EK'=CK', \angle{E'EK'}=\angle{BCK'}\\ \shoveleft \implies \angle{K'CN'}=300^{\circ}-\angle{BCK'}-\angle{ACN'}=120^{\circ}+\angle{CAG}-\angle{E'EK'}\\ \shoveleft = \angle{BAF}-\angle{E'EK'}=\angle{E'EM'}-\angle{E'EK'}=\angle{K'EM'}\\ \shoveleft CN'=AG=AF=EM' \implies \triangle{M'EK'}\cong\triangle{N'CK'}\\ \shoveleft \implies M'K'=N'K',\angle{EK'M'}=\angle{CK'N'} \implies \triangle{K'M'N'} \text{ is equilateral}\\ \shoveleft \implies \triangle{KMN} \text{ is equilateral triangle}\blacksquare \end{multline}\)