If two triangles have proportional sides then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal. * Let $ABC$ and $DEF$ be two triangles having proportional sides, (so that) as $AB$ (is) to $BC$, so $DE$ (is) to $EF$, and as $BC$ (is) to $CA$, so $EF$ (is) to $FD$, and, further, as $BA$ (is) to $AC$, so $ED$ (is) to $DF$. * I say that triangle $ABC$ is equiangular to triangle $DEF$, and (that the triangles) will have the angles which corresponding sides subtend equal. * (That is), (angle) $ABC$ (equal) to $DEF$, $BCA$ to $EFD$, and, further, $BAC$ to $EDF$.
Similar triangles are equiangular (this is the converse of Prop. 6.04).
Proofs: 1
