On sides $AB$ and $BC$ of an equilateral triangle $ABC$ are taken points $D$ and $K$, and on the side $AC$ , points $E$ and $M$ so that $DA + AE = KC +CM = AB$. Prove that the angle between lines $DM$ and $KE$ is equal to $60^o$.
Source:
Tags: geometry, angles, Equilateral
On sides $AB$ and $BC$ of an equilateral triangle $ABC$ are taken points $D$ and $K$, and on the side $AC$ , points $E$ and $M$ so that $DA + AE = KC +CM = AB$. Prove that the angle between lines $DM$ and $KE$ is equal to $60^o$.