Geometrical Problems Deducible from the First Six Books of Euclid, Arranged and Solved: To which is Added an Appendix Containing the Elements of Plane Trigonometry ...J. Smith, 1821 - 438 σελίδες |
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Αποτελέσματα 1 - 5 από τα 59.
Σελίδα 1
... shewn to be greater than AC ; there- fore AC is the least . ( 2. ) If a perpendicular be drawn bisecting a given straight line ; any point in this perpendicular is at equal A distances , and any point without the perpendicular is at THE ...
... shewn to be greater than AC ; there- fore AC is the least . ( 2. ) If a perpendicular be drawn bisecting a given straight line ; any point in this perpendicular is at equal A distances , and any point without the perpendicular is at THE ...
Σελίδα 4
... shewn that Ap = p C. Hence AP and BP together are equal to BC , and Ap , pB are equal to Cp , pB . Now ( Eucl . i . 20. ) BC is less than Bp , PC , and therefore AP , PB are less than Ap , PB ; therefore , & c . ( 7. ) Of all straight ...
... shewn that Ap = p C. Hence AP and BP together are equal to BC , and Ap , pB are equal to Cp , pB . Now ( Eucl . i . 20. ) BC is less than Bp , PC , and therefore AP , PB are less than Ap , PB ; therefore , & c . ( 7. ) Of all straight ...
Σελίδα 5
... shewn that AH is greater than AG . And from A there can only be drawn to BC two straight lines equal to each other , viz . one on each side of AD . Make DE - DF , and join AE . Then AE = AF ( i . 2. ) . And besides AE no other line can ...
... shewn that AH is greater than AG . And from A there can only be drawn to BC two straight lines equal to each other , viz . one on each side of AD . Make DE - DF , and join AE . Then AE = AF ( i . 2. ) . And besides AE no other line can ...
Σελίδα 11
... shewn that HI = IB ; and so on , if there be any other parts ; therefore AG , GH , HI , IB , & c . are all equal , and AB is divided as was required . COR . If it be required to divide the line into parts which shall have a given ratio ...
... shewn that HI = IB ; and so on , if there be any other parts ; therefore AG , GH , HI , IB , & c . are all equal , and AB is divided as was required . COR . If it be required to divide the line into parts which shall have a given ratio ...
Σελίδα 32
... shewn that every line drawn from A to BCD will be divided by the circum- ference of the circle GFH in the same ratio , i . e . GFH will be the locus required . ( 14. ) Having given the radius of a circle ; to de- termine its centre ...
... shewn that every line drawn from A to BCD will be divided by the circum- ference of the circle GFH in the same ratio , i . e . GFH will be the locus required . ( 14. ) Having given the radius of a circle ; to de- termine its centre ...
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Άλλες εκδόσεις - Προβολή όλων
Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD angle ABC angle equal BC is equal centre chord circle cutting circumference construct the triangle cosecant cosine describe a circle divided equation equiangular Eucl extremities find the value given angle given circle given difference given in position given line given point given ratio given rectangle given square given straight line Hence inscribed intercepted isosceles triangle Join AE least common multiple Let AB Let ABC let fall line given line joining lines be drawn lines drawn mean proportional meeting opposite side parallel to AC parallelogram pendicular point of intersection produced quadrant radius rectangle contained right angles right-angled triangle segments semicircle shewn sine squares of AC tang tangent transposition trapezium triangle ABC triangle required vertex vertical angle whence
Δημοφιλή αποσπάσματα
Σελίδα i - IF a straight line be divided into two equal, and also into two unequal parts ; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.
Σελίδα xi - IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle,. shall be equal to the square of the line which touches it.
Σελίδα 319 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.
Σελίδα 150 - Iff a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other...
Σελίδα 204 - FC are equal to one another : wherefore the circle described from the centre F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC.
Σελίδα 115 - If from a point, without a parallelogram, there be drawn two straight lines to the extremities of the two opposite sides, between which, when produced, the point does not lie, the difference of the triangles thus formed is equal to half the parallelogram. Ex. 2. The two triangles, formed by drawing straight lines from any point within a parallelogram to the extremities of its opposite sides, are together half of the parallelogram.
Σελίδα 14 - In one of the given equations obtain the value of one of the unknown quantities in terms of the other unknown quantity; Substitute this value in the other equation and solve.
Σελίδα 291 - AB describe a segment of a circle containing an angle equal to the given angle, (in.
Σελίδα 297 - Given the vertical angle, the difference of the two sides containing it, and the difference of the segments of the base made by a perpendicular from the vertex ; construct the triangle.
Σελίδα 90 - If from any point in the base of an isosceles triangle perpendiculars...