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, 1827 - 377 σελίδες |
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Αποτελέσματα 1 - 5 από τα 61.
Σελίδα 1
... shewn to be greater than AC ; there- fore AC is the least . given ( 2. ) If a perpendicular be drawn bisecting straight line ; any point in this perpendicular is at equal A distances , and any point without the perpendicular is at.
... shewn to be greater than AC ; there- fore AC is the least . given ( 2. ) If a perpendicular be drawn bisecting straight line ; any point in this perpendicular is at equal A distances , and any point without the perpendicular is at.
Σελίδα 4
... shewn that Ap = p C. Hence AP and BP together are equal to BC , and Ap , B are equal to Cp , p B. Now ( Eucl . i . 20. ) BC is less than Bp , pC , and therefore AP , PB are less than Ap , pB ; therefore , & c . Ρ ( 7. ) Of all straight ...
... shewn that Ap = p C. Hence AP and BP together are equal to BC , and Ap , B are equal to Cp , p B. Now ( Eucl . i . 20. ) BC is less than Bp , pC , and therefore AP , PB are less than Ap , pB ; therefore , & c . Ρ ( 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
Geometrical Problems Deducible from the First Six Books of Euclid: Arranged ... Miles Bland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2015 |
Geometrical Problems Deducible From the First Six Books of Euclid: Arranged ... Miles Bland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2023 |
Geometrical Problems Deducible from the First Six Books of Euclid: Arranged ... Miles Bland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2018 |
Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD angle ABC base bisect the angle centre chord circle ABC circles cut circumference describe a circle divided draw a line drawn parallel duplicate ratio equal angles equiangular Eucl extremities G draw given angle given circle given in position given line given point given ratio given square given straight line intercepted isosceles triangle Join AB Join AE Join BD Let AB Let ABC let fall line given line joining line required lines be drawn lines drawn mean proportional opposite sides parallel to AC parallelogram pendicular point of bisection point of contact point of intersection radius rectangle rectangle contained right angles right-angled triangle segments semicircle shewn tangent touching the circle trapezium triangle ABC whence
Δημοφιλή αποσπάσματα
Σελίδα 10 - 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 xv - 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 xxx - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Σελίδα 303 - 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 140 - 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 329 - CE is equal to the difference of the segments of the base made by the perpendicular.
Σελίδα 109 - 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 164 - PROPOSITION I. PROBLEM. — To describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it.
Σελίδα 281 - 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 270 - AB describe a segment of a circle containing an angle equal to the given angle, (in.