Manual of Euclid. Books IV. V. VI.1859 |
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Página 5
... perpendicular to any side BC ; the circle described from the centre D with the radius DF is inscribed in the given tri- angle . Draw DE and DG perpendicular to BA and AC . - B F Proof . In the triangles DEB , DFB , the angles DEB and ...
... perpendicular to any side BC ; the circle described from the centre D with the radius DF is inscribed in the given tri- angle . Draw DE and DG perpendicular to BA and AC . - B F Proof . In the triangles DEB , DFB , the angles DEB and ...
Página 7
... perpendicular to AB and AC ; and from their point of concourse F draw FA , FB , and FC ; B the circle described from the centre F with the radius FA is circumscribed about the given triangle . Proof . In the triangles FDA , FDB , the ...
... perpendicular to AB and AC ; and from their point of concourse F draw FA , FB , and FC ; B the circle described from the centre F with the radius FA is circumscribed about the given triangle . Proof . In the triangles FDA , FDB , the ...
Página 9
... perpendicular to it , and join AB , BC , CD , DA . ABCD is a square inscribed in the given circle . Proof . Because the angles at E are right , and therefore equal , the arcs on which they stand are equal , and therefore their chords ...
... perpendicular to it , and join AB , BC , CD , DA . ABCD is a square inscribed in the given circle . Proof . Because the angles at E are right , and therefore equal , the arcs on which they stand are equal , and therefore their chords ...
Página 67
... perpendicular let fall from its vertex to its base . PROPOSITION I. - THEOREM . Πρότασις α ' .— Τὰ τρίγωνα καὶ τὰ παραλληλόγραμμα , τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα , πρὸς ἄλληλὰ ἐστιν ὡς αἱ βάσεις . Triangles and parallelograms of the same ...
... perpendicular let fall from its vertex to its base . PROPOSITION I. - THEOREM . Πρότασις α ' .— Τὰ τρίγωνα καὶ τὰ παραλληλόγραμμα , τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα , πρὸς ἄλληλὰ ἐστιν ὡς αἱ βάσεις . Triangles and parallelograms of the same ...
Página 69
... perpendicular drawn from the point E to AB , and they are to one another as their bases . For the same reason , the triangle CDE is to the triangle ADE , as CE to EA . Therefore , as BD is to DA , so is CE to EA ( V. 11 ) . Statement ...
... perpendicular drawn from the point E to AB , and they are to one another as their bases . For the same reason , the triangle CDE is to the triangle ADE , as CE to EA . Therefore , as BD is to DA , so is CE to EA ( V. 11 ) . Statement ...
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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 Annotation arc BC base BC is equal bisected centre circumscribed circle Corollary diameter draw drawn equal angles equal to F equi equiangular equilateral equimultiples exscribed circles find the locus fore fourth given circle given lines given point given ratio given straight lines given triangle greater ratio harmonically homologous sides inscribed intersection joining less mean proportional multiple parallel parallelogram perpendicular polygon Proof Prop Q. E. D. PROPOSITION quadrilateral radical axis radius rectangle rectilinear figure remaining angle right angle right line segments similar square Statement.-Let tangents tiple triangle ABC vertex vertical angle Wherefore αἱ ἀνάλογον δὲ δοθέντα ἐν ἔσται ἴσα ἰσάκις ἴσον καὶ κύκλον λόγον μεγέθη περὶ πλευραὶ πρὸς τὸ Πρότασις πρῶτον τὰ τὰς τὴν τῆς τοῖς τὸν αὐτὸν τοῦ τρίγωνα τῷ τῶν
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 78 - 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.