|BOOK XI. 20.||241|
If they are not all equal, let BAC be that angle which is not less than either of the other two, and is greater than one of them, BAD.
At the point A in the straight line BA, make, in the plane which passes
through BA, AC, the angle BAE equal to the angle BAD; [I.23. make AE equal to AD; [I. 3. through E draw BEC cutting AB,AC at the points B,C; and join DB,DC.
Then, because AD is equal to AE, [Construction. and AB is. common to the two triangles BAD, BAE, the two sides BA, AD are equal to the two sides BA, AE, each to each ; and the angle BAD is equal to the angle BAE; [Constr. therefore the base BD is equal to the base BE. [I. 4.
And because BD, DC are together greater than BC, [1. 20. and one of them BD has been shewn equal to BE a part of EC. therefore the other DC is greater than the remaining part EC.
And because AD is equal to AE, [Construction. and AC is common to the two triangles DAC, EAC, but the base DC is greater than the base EC; therefore the angle DAC is greater than the angle EAC. [I. 25. And, by construction, the angle BAD is equal to the angle BAE ; therefore the angles BAD, DAC are together greater than the angles BAE, EAC, that is, than the angle BAC.
But the angle BAC is not less than either of the angles BAD, DAC;
therefore the angle BAC together with either of the other angles is greater than the third.
Wherefore, if a solid angle &c. q.e.d.