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, 1819 - 377 sider |
Fra bogen
Resultater 1-5 af 59
Side 4
... shewn that Ap = pC . 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 = pC . 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 ...
Side 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 be ...
... 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 be ...
Side 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 ...
Side 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 ...
Side 33
... shewn that BO = CO ; and the three lines OA , OB , OC being equal , a circle described from the centre O at the distance of any one of them will pass through the extremities of the other two . And besides this , no other circle can pass ...
... shewn that BO = CO ; and the three lines OA , OB , OC being equal , a circle described from the centre O at the distance of any one of them will pass through the extremities of the other two . And besides this , no other circle can pass ...
Indhold
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Andre udgaver - Se alle
Almindelige termer og sætninger
ABCD angle ABC angles at F base centre chord circle ABC circles cut circles touch circumference describe a circle divided draw a line draw the diameter 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 given triangle intercepted isosceles triangle Join AE Join BD Let AB Let ABC let fall line given line joining line required lines be drawn lines drawn mean proportional opposite side parallel to BC parallelogram pendicular point of bisection point of contact point of intersection point required radius rectangle right angles segments semicircle shewn tangent touching the circle trapezium triangle ABC
Populære passager
Side 14 - 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.
Side xiii - 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.
Side 230 - To describe an isosceles triangle, having each of the angles at the base double of the third angle.
Side 327 - 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.
Side 158 - 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...
Side 212 - 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.
Side 123 - 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.
Side 305 - 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.
Side 247 - The perpendicular from the vertex on the base of an equilateral triangle is equal to the side of an equilateral triangle inscribed in a circle, whose diameter is the base.
Side 299 - AB be equal to the given bisecting line ; and upon it describe a segment of a circle containing an angle equal to the given angle.