Bibliotheca Polyglotta

PROPOSITION 5.

PROBL. 5. PROPOS. 5.

第五題

About a given triangle to circumscribe a circle.

CIRCA datum triangulum circulum describere.

三角形。求作形外切圜。

Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC.

SIT circulus describendus circa datum triangulum ABC.

法曰。甲乙丙角形。求作形外切圜。

Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC.

Diuidanturduo latera AB, AC, (quae in triangulo rectangulo, vel obtusangulo sumenda sunt facilitatis gratia, circa rectum, vel obtusum angulum, quamvis hoc non sit omnino necessarium, sed duo quaeuis latera bifariam possint secari) bifariam in D, et E, punctis, ex quibus educantur DF, EF, perpendiculares ad dicta latera, coeuntes in F, (Quod enim coeaht, patet. Nam si ducta esset recta DE, fierent anguli FDE, FED, duobus rectis minores.) eritque F, vel intra triangulum, vel in latere BC, vel extra triangulum.

先平分兩邊。若形是直角、鈍角。則分直角鈍角、之兩旁邊。於丁於戊。次於丁、戊、上各作垂線。為己丁、己戊。而相遇於己。若自丁至戊、作直線。卽己丁戊角形之己丁戊、己戊丁、兩角。小於兩直角。故丁己、戊己、兩線必相遇。其己點或在形內。或在形外。

First let them meet within at F and let FB, FC, FA be joined. Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4] Similarly we can prove that CF is also equal to AF; so that FB is also equal to FC; therefore the three straight lines FA, FB, FC are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle ABC. Let it be circumscribed, as ABC.

Ducantur rectae FA, FB, FC. Quoniam igitur latera AD, DF, trianguli ADF, aequalia sunt lateribus BD, DF, trianguli BDF, et anguli ad D, recti; erunt bases FA, FB, aequales. ¹ Eodem modo erunt FA, FC, aequales. Cum ergo tres rectae FA, FB, FC, sint aequales, circulus descriprus ex F, ad interuallum FA, transibit quoque per puncta B, et C.

1. 4. primi.

俱作己甲、己乙、己丙、三線。或在乙丙邊上。止作己甲線。其甲丁己角形之甲丁。與乙丁己角形之乙丁。兩腰等。丁己同腰。而丁之兩旁角、俱直角。卽甲己、己乙、兩底必等。一卷四依顯甲己戊、丙己戊、兩形之甲己、己丙、兩底亦等。則己甲、己乙、己丙、三線俱等。末作圜。以己為心。甲為界。必切丙乙、而為角形之形外切圜。

Next, let DF, EF meet on the straight line BC at F, as is the case in the second figure; and let AF be joined. Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC.

Again, let DF, EF meet outside the triangle ABC at F, as is the case in the third figure, and let AF, BF, CF be joined. Then again, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I. 4] Similarly we can prove that CF is also equal to AF; so that BF is also equal to FC; therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and will have been circumscribed about the triangle ABC.

Therefore about the given triangle a circle has been circumscribed. Q. E. F.

Circa datum ergo triangulum circulum descripsimus. Quod erat faciendum.

And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31]

HINC manifestum est, si centrum intra triangulum cadat,² omnes angulos esse acutos, quoniam omnes sunt in maiori segmento circuli: si vero sit in latere BC,³ angulum BAC, ei lateri oppositum, esse rectum, quod sit in semicirculo: si denique cadat extra triangulum,⁴ angulum oppositum BAC, obtusum esse, cum sit in minori segmento circuli.

2. 31. tertij. 3. 31. tertij. 4. 31. tertij.

一系。若圜心在三角形內。卽三角形為銳角形。何者。每角在圜大分之上故。若在一邊之上。卽為直角形。若在形外。卽為鈍角形。

CONTRA vero perspicuum est, si triangulum fuerit acutangulum, cadere intra triangulum: si rectangulum, in latus recto angulo oppositum: si denique obtusangulum fuerit, extra triangulum. Quod quidem facile ostendetur, ducendo ad incommodum aliquod, sive absurdum. Quiasi in acutangulo caderet centrum in vnum latus, esset angulus ei oppofitus rectus: si vero extra, esset idem angulus obtusus. Item si in rectangulo centrum caderet intra, essent omnes anguli acuti, si vero extra, esset angulus oppositus obtusus. Denique si in triangulo obtusangulo caderet in vnum latus, esset angulus ei oppositus, rectus, si vero intra, omnes anguli essent acuti. Quae omnia ex priori parte huius corollarium colliguntur, et pugnant cum hypothesi.

二系。若三角形為銳角形。卽圜心必在形內。若直角形。必在一邊之上。若鈍角形。必在形外。

COLLIGITVR etiam ex hoc problemate, quanam arte describendus sit circulus, qui per duta tria puncta non in vna recta linea existentia transeat. Nam si data puncta tribus rectis iungantur, ut constituatur triangulum, facile ciica ipsum circulus describetur, ut hac propositione traditum est. Quod tamen facilius efficietur praexiilla, quam tradidimus propos.25.lib.3. Sint enim datis tria puncta A, B, C; Ex A, et B, quouis interuallo eodem duo arcus describantur se intersecantes in D, et E, punctis, per quae recta linea ducatur DH. Item ex A, et C, quouis alio interuallo eodem, vel etiam, si placet, priori illo, alij duo arcus delineentur secantes sese in F, et G, punctis, per quae recta ducatur FH, secans rectam DH, in H. Dico H, esse centrum circuli transeuntis per data puncta A, B, et C. Nam si ducerentur rectae AB, AC, BC, diuiderentur latera AB, AC, trianguli ABC, bifariam a rectis DH, FH, ceu demonstratum est in praxi illa propos.25.lib.3. Quare ut in hoc 5. problemate Euclides ostendit, H, erit centrum circuli circa triangulum ABC, descripti. Quod est propositum.

增。從此推得一法。任設三點。不在一直線。可作一過三點之圜。其法、先以三點作三直線相聯。成三角形。次依前作。
其用法。甲、乙、丙、三點。先以甲、乙、兩點各自為心。相向、各任作圜分。令兩圜分相交於丁、於戊。次甲、丙、兩點。亦如之。令兩圜分相交於己、於庚。末作丁戊、己庚、兩線、各引長之。令相交於辛。卽辛為圜之心。論見三卷二十五增。

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