Therefore the triangle ABC is equiangular with the triangle DGF.
Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4]
But, by hypothesis, as BA is to AC, so also is ED to DF;
therefore also, as ED is to DF, so is GD to DF. [V. 11]
Therefore ED is equal to DG;
[V. 9] and DF is common;
therefore the two sides ED, DF are equal to the two sides GD, DF;
and the angle EDF is equal to the angle GDF;
therefore the base EF is equal to the base GF,
and the triangle DEF is equal to the triangle DGF,
and the remaining angles will be equal to the remaining angles,
namely those which the equal sides subtend. [I. 4]
Therefore the angle DFG is equal to the angle DFE,
and the angle DGF to the angle DEF.
But the angle DFG is equal to the angle ACB;
therefore the angle ACB is also equal to the angle DFE.
And, by hypothesis, the angle BAC is also equal to the angle EDF;
therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32]
therefore the triangle ABC is equiangular with the triangle DEF.
Complete text
Book I