Book 1
21
Again, since, 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.
21
But BA, AC were proved greater than BE, EC; therefore BA, AC are much greater than BD, DC.
21
Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I. 16] therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED.
21
For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC. But the angle BDC was proved greater than the angle CEB; therefore the angle BDC is much greater than the angle BAC.
21
Therefore etc.
21
Q. E. D.
Proposition 22.
22
Enunciation Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [I. 20]
22
Proof. Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, and B, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C.
22
Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3]
22
With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined;
22
I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C.
22
For, since the point F is the centre of the circle DKL, FD is equal to FK.
22
But FD is equal to A; therefore KF is also equal to A.
22
Again, since the point G is the centre of the circle LKH, GH is equal to GK.
