Manual of Euclid. Books IV. V. VI.1859 |
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Página 4
... remaining angle D is equal to BAC ( I. 32 ) , and therefore the triangle BAC inscribed in the given circle is equiangular to the given triangle EDF . PROPOSITION III . - PROBLEM . Προτασις γ ́ . Περὶ τὸν δοθέντα κύκλον τῷ δοθέντι ...
... remaining angle D is equal to BAC ( I. 32 ) , and therefore the triangle BAC inscribed in the given circle is equiangular to the given triangle EDF . PROPOSITION III . - PROBLEM . Προτασις γ ́ . Περὶ τὸν δοθέντα κύκλον τῷ δοθέντι ...
Página 17
... AED , are equal , and therefore the sides FE and FA are equal ( I. 6 ) ; and in the same manner it is proved that the remaining lines FB , FC , and FD are equal ; C therefore the five lines FA , FB , FC , BOOK IV . 17.
... AED , are equal , and therefore the sides FE and FA are equal ( I. 6 ) ; and in the same manner it is proved that the remaining lines FB , FC , and FD are equal ; C therefore the five lines FA , FB , FC , BOOK IV . 17.
Página 21
... remaining lines as indicated in the figure . Proof . The points o , o ' , o " , o " are the centres of the circles . For , since Lo and LC are equal , the angles LoC and LCo are equal , and LoC is equal to the sum of oCA and oAC , or of ...
... remaining lines as indicated in the figure . Proof . The points o , o ' , o " , o " are the centres of the circles . For , since Lo and LC are equal , the angles LoC and LCo are equal , and LoC is equal to the sum of oCA and oAC , or of ...
Página 66
... remaining side of the se- cond is to the remaining side of the first . DEFINITION III . Ὅρος γ ́ . — Ακρον καὶ μέσον λόγον εὐθεῖα τετμῆσθαι λέγεται , ὅταν ἦ ὡς ἡ ὅλη πρὸς τὸ μεῖζον τμήμα οὕτως τὸ μεῖζον πρὸς τὸ ἔλασσον . A straight line ...
... remaining side of the se- cond is to the remaining side of the first . DEFINITION III . Ὅρος γ ́ . — Ακρον καὶ μέσον λόγον εὐθεῖα τετμῆσθαι λέγεται , ὅταν ἦ ὡς ἡ ὅλη πρὸς τὸ μεῖζον τμήμα οὕτως τὸ μεῖζον πρὸς τὸ ἔλασσον . A straight line ...
Página 74
Euclid Joseph Allen Galbraith. logous to ED , and in the triangle DEG the remaining angle G is equal to the angle C in the triangle ABČ . ( I. 32 ) . Proof . Because the triangles ABC and DEG are equiangular , BA is to AC as ED to DG ...
Euclid Joseph Allen Galbraith. logous to ED , and in the triangle DEG the remaining angle G is equal to the angle C in the triangle ABČ . ( I. 32 ) . Proof . Because the triangles ABC and DEG are equiangular , BA is to AC as ED to DG ...
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Manual of Euclid: Books IV., V., VI., by J.A. Galbraith and S. Haughton Euclides Sin vista previa disponible - 2015 |
Términos y frases comunes
angle BAC angles ABC arc BC bisected centre circumscribed circle construct the triangle Corollary diameter draw drawn equal angles equiangular equilateral equimultiples exscribed circles find the locus fore fourth given circle given in position given line given point given ratio given triangle greater ratio homologous homologous sides inscribed intersection joining less multiple parallel parallelogram pentagon perpendicular polygon Proof Proof.-Because Prop Q. E. D. Annotation Q. E. D. PROPOSITION quadrilateral radical axis radius rectangle rectilinear figure right angles right line scribed segments square Statement.-Let straight line tangents third tiple triangle ABC vertex vertical angle Wherefore αἱ ἀνάλογον γωνίας δὲ δοθέντι ἐν ἔσται ἐστὶ ἴσα ἰσάκις ἴσας ἴσον καὶ λόγον μεγέθη Ὅρος ὅταν περὶ πρὸς τὸ Πρότασις τὰ τὰς τε καὶ τὴν τῆς τοῖς τὸν αὐτὸν τοῦ τῷ τῶν ὡς
Pasajes populares
Página 38 - If the first be the same multiple of the second which the third is of the fourth, and if of the first and third there be taken equimultiples, these shall be equimultiples, the one of the second, and the other of the fourth. Let A the first be the same multiple of B the second, that C the third is of D the fourth ; and of A...
Página 28 - The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth ; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth: or, if the multiple of the first be equal to that of the second, the multiple of the third is also equal...
Página 3 - IN a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle ; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.
Página 70 - DE ; but equal triangles on the same base and on the same side of it, are between the same parallels ; (i.
Página 92 - CF ; but K has to M the ratio which is compounded of the ratios of the sides ; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. Wherefore equiangular parallelograms, &c.
Página 71 - Now let BD be to DC, as BA to AC, and join AD ; the angle CAD is equal to the angle DAE. The same construction being made, because BD is to DC as BA to AC ; and also BD to DC, BA to AF (2.
Página 29 - When of the equimultiples of four magnitudes (taken as in the fifth definition), the multiple of the first is greater than that of the second, but the multiple of the third is not greater than the multiple of the fourth ; then the first is said to have to the second a greater ratio than the third magnitude has to the fourth...
Página 60 - D, and other four E, F, G, H, which, two and two, have the same ratio, viz. as A is to B, so is E to F; and as B...
Página 66 - A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.
Página 68 - CF; and because it has been shewn, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF.