298260Apollonij Pergæi
345[Figure 345]
las A O, K H ſunt anguli alterni A O K, &
H K O æquales inter ſe;
igitur
angulus A O K æqualis erit angulo C A H; & propterea in duobus triangulis
K A O, & H C A tertius angulus A C H æqualis erit tertio angulo K A O,
& propterea triangulum K A O iſoſcelium, & ſimile erit triangulo H A C,
ſiuè F G E; igitur conus, cuius vertex K baſis circulus A O perpendicularis
ad planum trianguli A K O erit conus rectus, & ſimilis cono E F G dato.
angulus A O K æqualis erit angulo C A H; & propterea in duobus triangulis
K A O, & H C A tertius angulus A C H æqualis erit tertio angulo K A O,
& propterea triangulum K A O iſoſcelium, & ſimile erit triangulo H A C,
ſiuè F G E; igitur conus, cuius vertex K baſis circulus A O perpendicularis
ad planum trianguli A K O erit conus rectus, & ſimilis cono E F G dato.
Alioquin contineat illum conus alius, cuius vertex ſit Q, &
triangu-
11d lum Q A P, & oſtendetur quemadmodum dictum eſt, quod planum
tranſiens per axim illius coni erectum ad planum ſectionis A B C ſectio
communis cum plano ſectionis eſt A C, & quod punctum verticis illius
coni ſit in circumferentia ſegmenti A H C, & c. Quia ſupponitur, quod
conus Q A P ſimilis cono E F G contineat ellipſim A B C, cuius axis tranſuer-
ſus C A, & latus rectum A D; igitur triangulum per axim coni ductum Q
A P, nedum ſimile erit triangulo E F G, ſed etiam perpendiculare erit ad pla-
num ellipſis A B C, & propterea conſiſtet in plano circularis ſegmenti A H C
pariter erecti ad planum A B C, per idem axim A C extenſum, & eſt angu-
lus A Q C æqualis angulo verticali F propter ſimilitudinem duorum triangu-
lorum, & ex conſtructione primæ partis huius propoſitionis, eſt ſegmentum A
H C capax anguli æqualis angulo F; ſecaturque bifariam in H; igitur angulus
A Q C æqualis ipſi F in peripheria ſegmenti A H C exiſtit. Ducatur poſtea
Q S parallela lateri tranſuer ſo ellipſis A C, quæ ſecet baſim trianguli per axim
Q A P productam in S, & à puncto H bipartitæ diuiſionis ſegmenti A H C
coniungatur recta linea H Q producaturq; quouſq; occurratrectæ lineæ C A in R.
Quoniã duo anguli A H C, & A Q C in eodẽ circuli ſegmento conſtituti æqua-
les ſunt inter ſe; pariterq; duo anguli C A H, & C Q H in eodẽ circuli ſegmento
exiſtentes ſunt æquales, & eſt angulus A P Q æqualis angulo P A Q in triangu-
lo iſoſcelio Q A P; & angulus P A Q æqualis angulo C A H in triangulis ſimi-
libus; igitur angulus A P Q æqualis eſt alterno angulo P Q H; &
11d lum Q A P, & oſtendetur quemadmodum dictum eſt, quod planum
tranſiens per axim illius coni erectum ad planum ſectionis A B C ſectio
communis cum plano ſectionis eſt A C, & quod punctum verticis illius
coni ſit in circumferentia ſegmenti A H C, & c. Quia ſupponitur, quod
conus Q A P ſimilis cono E F G contineat ellipſim A B C, cuius axis tranſuer-
ſus C A, & latus rectum A D; igitur triangulum per axim coni ductum Q
A P, nedum ſimile erit triangulo E F G, ſed etiam perpendiculare erit ad pla-
num ellipſis A B C, & propterea conſiſtet in plano circularis ſegmenti A H C
pariter erecti ad planum A B C, per idem axim A C extenſum, & eſt angu-
lus A Q C æqualis angulo verticali F propter ſimilitudinem duorum triangu-
lorum, & ex conſtructione primæ partis huius propoſitionis, eſt ſegmentum A
H C capax anguli æqualis angulo F; ſecaturque bifariam in H; igitur angulus
A Q C æqualis ipſi F in peripheria ſegmenti A H C exiſtit. Ducatur poſtea
Q S parallela lateri tranſuer ſo ellipſis A C, quæ ſecet baſim trianguli per axim
Q A P productam in S, & à puncto H bipartitæ diuiſionis ſegmenti A H C
coniungatur recta linea H Q producaturq; quouſq; occurratrectæ lineæ C A in R.
Quoniã duo anguli A H C, & A Q C in eodẽ circuli ſegmento conſtituti æqua-
les ſunt inter ſe; pariterq; duo anguli C A H, & C Q H in eodẽ circuli ſegmento
exiſtentes ſunt æquales, & eſt angulus A P Q æqualis angulo P A Q in triangu-
lo iſoſcelio Q A P; & angulus P A Q æqualis angulo C A H in triangulis ſimi-
libus; igitur angulus A P Q æqualis eſt alterno angulo P Q H; &