## 191.II.CONSTRUCTIO PROBLEMATIS OPTICI. Propoſitio 39 Libri v. Alhazeni, & 22 lib. VI . Vitellionis.

Puncta B, C, & circulus E K, cujus centrum eſt A,
data ſunt in eodem plano; inveniendum eſt pun-
ctum K in peripheria circuli, ita ut lineæ B K, C K
faciant cum linea A K angulos inter ſe æquales.

### 191.1.

TAB. LVI.
fig. 1.

DUctis A B, A C; fiat A C, A F: : A F, A Q; & A B,
A E: : A E, A P: ſint etiam A P, & A Q, bifariam di-
viſæ in R & S; in angulo B A C ſint perſecta parallelo-
gramma P A Q H & A R Z S. In R Z producta ſumantur
Z Y & Z X, utraque æqualis lineæ quæ poteſt differentiam
inter quadrata Q S & Z S: fiat X V æqualis X Y & paral-
lela A B; & lateribus X V & X Y deſcribatur hyperbola,
quæ tranſibit per puncta Q & H, uti patet per conſtructio-
nem; hyperbola hæc Q X H occurret circulo in puncto K,
quod quæritur.

Ductis K O & KI parallelis A C & A B, quarum K I
occurrit Y X in puncto D. Ob hyperbolam rectangulum
Y D X æquale eſt quadrato K D ordinatæ, vel quadrato O R; & rectangulum Y T X æquale eſt quadrato H T, vel quadra-
to P R; & demto a rectangulo Y T X, rectangulum Y D X
lum R D T vel A I Q, quod æquale erit rectangulo A O P. P O ergo eſt ad A I, vel O K ipſi æqualem, ut Q I ad
A O, vel I K; & ductis lineis K P, K Q triangula K O P,
K I Q erunt ſimilia & ideo æquiangula; idcirco anguli
A P K, A Q K, qui iidem ſunt, vel ſupplementa angulorum
æqualium O P K, I Q K erunt inter ſe æquales; ſed per con-
ſtructionem A B, A E vel A K, : : A K vel A E, A P; ideo duo triangula B A K, K A P ſunt ſimilia, & ob ean-
dem cauſam duo triangula C A K, K A Q, ſunt etiam ſi-
milia; ideo angulus B K A eſt æqualis angulo A P K, &

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