--- book: 11 number: 23 id: "XI.23" kind: "construction" uses: ["[[book-11/proposition-22]]", "[[book-1/proposition-8]]", "[[book-6/proposition-2]]", "[[book-1/proposition-29]]", "[[book-6/proposition-4]]", "[[book-5/proposition-16]]", "[[book-1/proposition-25]]", "[[book-11/proposition-12]]", "[[book-1/proposition-4]]", "[[book-1/proposition-47]]", "[[book-4/proposition-1]]", "[[book-3/proposition-31]]"] source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.23" license: "CC-BY-SA-4.0" --- # XI.23 *To construct a solid angle out of three plane angles two of which*, *taken together in any manner*, *are greater than the remaining one: thus the three angles must be less than four right angles.* ## Proof Let the angles *ABC*, *DEF*, *GHK* be the three given plane angles, and let two of these, taken together in any manner, be greater than the remaining one, while, further, the three are less than four right angles; thus it is required to construct a solid angle out of angles equal to the angles *ABC*, *DEF*, *GHK*. Let *AB*, *BC*, *DE*, *EF*, *GH*, *HK* be cut off equal to one another, and let *AC*, *DF*, *GK* be joined; it is therefore possible to construct a triangle out of straight lines equal to *AC*, *DF*, *GK*. [[book-11/proposition-22|XI. 22]] Let *LMN* be so constructed that *AC* is equal to *LM*, *DF* to *MN*, and further *GK* to *NL*, let the circle *LMN* be described about the triangle *LMN*, let its centre be taken, and let it be *O*; let *LO*, *MO*, *NO* be joined; I say that *AB* is greater than *LO*. For, if not, *AB* is either equal to *LO*, or less. First, let it be equal. Then, since *AB* is equal to *LO*, while *AB* is equal to *BC*, and *OL* to *OM*, the two sides *AB*, *BC* are equal to the two sides *LO*, *OM* respectively; and, by hypothesis, the base *AC* is equal to the base *LM*; therefore the angle *ABC* is equal to the angle *LOM*. [[book-1/proposition-8|I. 8]] For the same reason the angle *DEF* is also equal to the angle *MON*, and further the angle *GHK* to the angle *NOL*; therefore the three angles *ABC*, *DEF*, *GHK* are equal to the three angles *LOM*, *MON*, *NOL*. But the three angles *LOM*, *MON*, *NOL* are equal to four right angles; therefore the angles *ABC*, *DEF*, *GHK* are equal to four right angles. But they are also, by hypothesis, less than four right angles: which is absurd. Therefore *AB* is not equal to *LO*. I say next that neither is *AB* less than *LO*. For, if possible, let it be so, and let *OP* be made equal to *AB*, and *OQ* equal to *BC*, and let *PQ* be joined. Then, since *AB* is equal to *BC*, *OP* is also equal to *OQ*, so that the remainder *LP* is equal to *QM*. Therefore *LM* is parallel to *PQ*, [[book-6/proposition-2|VI. 2]] and *LMO* is equiangular with *PQO*; [[book-1/proposition-29|I. 29]] therefore, as *OL* is to *LM*, so is *OP* to *PQ*; [[book-6/proposition-4|VI. 4]] and alternately, as *LO* is to *OP*, so is *LM* to *PQ*. [[book-5/proposition-16|V. 16]] But *LO* is greater than *OP*; therefore *LM* is also greater than *PQ*. But *LM* was made equal to *AC*; therefore *AC* is also greater than *PQ*. Since, then, the two sides *AB*, *BC* are equal to the two sides *PO*, *OQ*, and the base *AC* is greater than the base *PQ*, therefore the angle *ABC* is greater than the angle *POQ*. [[book-1/proposition-25|I. 25]] Similarly we can prove that the angle *DEF* is also greater than the angle *MON*, and the angle *GHK* greater than the angle *NOL*. Therefore the three angles *ABC*, *DEF*, *GHK* are greater than the three angles *LOM*, *MON*, *NOL*. But, by hypothesis, the angles *ABC*, *DEF*, *GHK* are less than four right angles; therefore the angles *LOM*, *MON*, *NOL* are much less than four right angles. But they are also equal to four right angles: which is absurd. Therefore *AB* is not less than *LO*. And it was proved that neither is it equal; therefore *AB* is greater than *LO*. Let then *OR* be set up from the point *O* at right angles to the plane of the circle *LMN*, [[book-11/proposition-12|XI. 12]] and let the square on *OR* be equal to that area by which the square on *AB* is greater than the square on *LO*; [Lemma] let *RL*, *RM*, *RN* be joined. Then, since *RO* is at right angles to the plane of the circle *LMN*, therefore *RO* is also at right angles to each of the straight lines *LO*, *MO*, *NO*. And, since *LO* is equal to *OM*, while *OR* is common and at right angles, therefore the base *RL* is equal to the base *RM*. [[book-1/proposition-4|I. 4]] For the same reason *RN* is also equal to each of the straight lines *RL*, *RM*; therefore the three straight lines *RL*, *RM*, *RN* are equal to one another. Next, since by hypothesis the square on *OR* is equal to that area by which the square on *AB* is greater than the square on *LO*, therefore the square on *AB* is equal to the squares on *LO*, *OR*. But the square on *LR* is equal to the squares on *LO*, *OR*, for the angle *LOR* is right; [[book-1/proposition-47|I. 47]] therefore the square on *AB* is equal to the square on *RL*; therefore *AB* is equal to *RL*. But each of the straight lines *BC*, *DE*, *EF*, *GH*, *HK* is equal to *AB*, while each of the straight lines *RM*, *RN* is equal to *RL*; therefore each of the straight lines *AB*, *BC*, *DE*, *EF*, *GH*, *HK* is equal to each of the straight lines *RL*, *RM*, *RN*. And, since the two sides *LR*, *RM* are equal to the two sides *AB*, *BC*, and the base *LM* is by hypothesis equal to the base *AC*, therefore the angle *LRM* is equal to the angle *ABC*. [[book-1/proposition-8|I. 8]] For the same reason the angle *MRN* is also equal to the angle *DEF*, and the angle *LRN* to the angle *GHK*. Therefore, out of the three plane angles *LRM*, *MRN*, *LRN*, which are equal to the three given angles *ABC*, *DEF*, *GHK*, the solid angle at *R* has been constructed, which is contained by the angles *LRM*, *MRN*, *LRN*. Q. E. F. Lemma. But how it is possible to take the square on *OR* equal to that area by which the square on *AB* is greater than the square on *LO*, we can show as follows. Let the straight lines *AB*, *LO* be set out, and let *AB* be the greater; let the semicircle *ABC* be described on *AB*, and into the semicircle *ABC* let *AC* be fitted equal to the straight line *LO*, not being greater than the diameter *AB*; [[book-4/proposition-1|IV. 1]] let *CB* be joined Since then the angle *ACB* is an angle in the semicircle *ACB*, therefore the angle *ACB* is right. [[book-3/proposition-31|III. 31]] Therefore the square on *AB* is equal to the squares on *AC*, *CB*. [[book-1/proposition-47|I. 47]] Hence the square on *AB* is greater than the square on *AC* by the square on *CB*. But *AC* is equal to *LO*. Therefore the square on *AB* is greater than the square on *LO* by the square on *CB*. If then we cut off *OR* equal to *BC*, the square on *AB* will be greater than the square on *LO* by the square on *OR*. Q. E. F.