If a solid angle is contained by three plane angles, then the sum of any two is greater than the remaining one.

Let the solid angle at *A* be contained by the three plane angles *BAC, CAD,* and *DAB.*

I say that the sum of any two of the angles *BAC, CAD,* and *DAB* is greater than the remaining one.

If the angles *BAC, CAD,* and *DAB* are equal to one another, then it is clear that the sum of any two is greater than the remaining one.

But, if not, let *BAC* be greater. In the plane through *BA* and *AC,* construct the angle *BAE* equal to the angle *DAB* at the point *A* on the straight line *AB.* Make *AE* equal to *AD,* draw *BEC* across through the point *E* cutting the straight lines *AB* and *AC* at the points *B* and *C,* and join *DB* and *DC.*

Now, since *DA* equals *AE,* and *AB* is common, therefore two sides are equal to two sides. And the angle *DAB* equals the angle *BAE,* therefore the base *DB* equals the base *BE.*

And, since the sum of the two sides *BD* and *DC* is greater than *BC,* and of these *DB* was proved equal to *BE,* therefore the remainder *DC* is greater than the remainder *EC.*

Now, since *DA* equals *AE,* and *AC* is common, and the base *DC* is greater than the base *EC,* therefore the angle *DAC* is greater than the angle *EAC.*

But the angle *BAE* equals the angle *DAB,* therefore the sum of the angles *DAB* and *DAC* is greater than the angle *BAC.*

Similarly we can prove that the sum of any two of the remaining angles is greater than the remaining one.

Therefore, *if a solid angle is contained by three plane angles, then the sum of any two is greater than the remaining one.*

Q.E.D.

Various interpretations have been made of the intent of the form of proof, but all require minor changes to clarify the structure.