Proposition I.21 · Theorem
If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.
Drag the points circled in red. The figure changes, and every step of the proof still holds.
- Let ABC be a triangle, and on BC, one of its sides, from its extremities let the two straight lines BD, DC be constructed meeting within the triangle at D. The claim: BD, DC are less than BA, AC, but contain a greater angle, the angle BDC greater than the angle BAC. (D can be dragged inside the triangle.)
- Let BD be drawn through to meet AC at E.Postulate 2Postulate 2: To produce a finite straight line continuously in a straight line.
- In the triangle ABE the two sides AB, AE are greater than BE. Let EC be added to each; therefore BA, AC are greater than BE, EC.Proposition I.20Proposition I.20: In any triangle two sides taken together in any manner are greater than the remaining one.
- Again, in the triangle CED the two sides CE, ED are greater than CD. Let DB be added to each; therefore CE, EB are greater than CD, DB.Proposition I.20Proposition I.20: In any triangle two sides taken together in any manner are greater than the remaining one.
- Therefore BA, AC are much greater than BD, DC.
- Again, since in the triangle CDE the exterior angle BDC is greater than the interior and opposite angle CED,Proposition I.16Proposition I.16: In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.
- and likewise in the triangle ABE the exterior angle CEB is greater than the interior and opposite angle BAE, the angle BDC is much greater than the angle BAC.Proposition I.16Proposition I.16: In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.
Being what it was required to prove.