Let polygon CAB and polygon FED be two triangles having the two sides line AB and line AC equal to the two sides line DE and line DF respectively, namely line AB equal to line DE and line AC equal to line DF, and the angle BAC equal to the angle EDF.

I say that the base line BC also equals the base line EF, the polygon CAB equals the polygon FED, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle angleABC equals the angle angleDEF, and the angle ACB equals the angle angleDFE.

If the polygon CAB is superposed on the polygon FED, and if the point A is placed on the point D and the straight line AB on line DE, then the point B also coincides with point E, because line AB equals line DE.

Again, line AB coinciding with line DE, the straight line AC also coincides with line DF, because the angle BAC equals the angle EDF. Hence, the point C also coincides with the point F, because line AC again equals line DF.

But point B also coincides with point E, hence the base line BC coincides with the base line EF and equals it.

Thus the whole polygon CAB coincides with the whole polygon FED and equals it.

And the remaining angles also coincide with the remaining angles and equal them, the angle angleABC equals the angle angleDEF, and the angle ACB equals the angle angleDFE.

Therefore if two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides.

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