Intermediate Geometry: Being Sections V and VI of "geometry, Theoretical and Practical".W.B. Clive, University Tutorial Press Ld., 1908 |
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Side 457
... concurrent straight lines . Prove that the locus of a point P which moves so that its perpendicular distance from OM is a mean proportional between its perpendicular distances from OL and ON is a pair of straight lines . 327. PROBLEM ...
... concurrent straight lines . Prove that the locus of a point P which moves so that its perpendicular distance from OM is a mean proportional between its perpendicular distances from OL and ON is a pair of straight lines . 327. PROBLEM ...
Side 469
... concurrent . 5. Find a fourth harmonic point to three given points . 6. Draw a fourth harmonic ray to three given rays . 7. A diameter of a circle is divided harmonically by any tangent and the orthogonal projection on the diameter of ...
... concurrent . 5. Find a fourth harmonic point to three given points . 6. Draw a fourth harmonic ray to three given rays . 7. A diameter of a circle is divided harmonically by any tangent and the orthogonal projection on the diameter of ...
Side 479
... concurrent . DEFINITION . - The radical centre of three circles is the point of intersection of their three radical axes . 351. THEOREM 3. The difference between the squares on the tangents from a given point to two given circles is ...
... concurrent . DEFINITION . - The radical centre of three circles is the point of intersection of their three radical axes . 351. THEOREM 3. The difference between the squares on the tangents from a given point to two given circles is ...
Side 485
... concurrent , then the product of the ratios of the segments into which these lines divide the opposite sides is equal to + 1 : and conversely . То prove BX CY AZ XC YA ZB 1. ( Figs . 491 , 492 , 493. ) Cons . - Through A draw a line ...
... concurrent , then the product of the ratios of the segments into which these lines divide the opposite sides is equal to + 1 : and conversely . То prove BX CY AZ XC YA ZB 1. ( Figs . 491 , 492 , 493. ) Cons . - Through A draw a line ...
Side 487
... concurrent and X , Y , Z ' are collinear , calculate ZZ ' ; given BX = 5 " , XC = 3 ′′ , CY = 3 ′′ , YA = 1 " , AB = 2 " . 3. Deduce from Ceva's Theorem that the following sets of lines connected with a triangle are concurrent : - ( a ) ...
... concurrent and X , Y , Z ' are collinear , calculate ZZ ' ; given BX = 5 " , XC = 3 ′′ , CY = 3 ′′ , YA = 1 " , AB = 2 " . 3. Deduce from Ceva's Theorem that the following sets of lines connected with a triangle are concurrent : - ( a ) ...
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Intermediate Geometry: Being Sections V and VI of Geometry, Theoretical and ... Walter Percy Workman Ingen forhåndsvisning tilgjengelig - 2023 |
Intermediate Geometry: Being Sections V and VI of Geometry, Theoretical and ... Walter Percy Workman Ingen forhåndsvisning tilgjengelig - 2023 |
Intermediate Geometry: Being Sections V and VI of Geometry, Theoretical and ... Walter Percy Workman Ingen forhåndsvisning tilgjengelig - 2019 |
Vanlige uttrykk og setninger
ABCD B.Sc base bisects centre of similitude Ceva's Theorem CHAPTER chord circles touch circumcircle collinear concurrent concyclic points Cons Cons.-Draw cyclic quadrilateral diagonals diameter dihedral angle divided equiangular EXERCISES externally figure Find the locus Geometry given circles given line given point given ratio given straight line harmonic range Hence hypotenuse intersecting planes irrational numbers LAOB LDPE lies in plane line drawn line joining line perpendicular mean proportional meet nine-point circle opposite edges orthocentre pair parallel planes parallelogram parallelopiped perpendicular Picture Plane plane X plane XX polygon Proof quadrilateral radical axis radii radius rational numbers rectangle contained remaining Theorems Required to prove respectively right angles segments similar triangles Similarly solid angle square tangents tetrahedron THEOREM.-If three lines triangle ABC Tutorial vertex vertical angle
Populære avsnitt
Side 570 - If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.
Side 584 - If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.
Side 575 - The square described on the hypothenuse of a rightangled triangle is equal to the sum of the squares described on the other two sides.
Side 424 - IN a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.
Side 10 - The historical part is concise and clear, but the criticism is even more valuable, and a number of illustrative extracts contribute a most useful feature to the volume.
Side 571 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 16 - The object of the UNIVERSITY TUTORIAL SERIES is to provide candidates for examinations and learners generally with text-books which shall convey in the simplest form sound instruction in accordance with the latest results of scholarship and scientific research. Important points are fully and clearly treated, and care has been taken not to introduce details which are likely to perplex the beginner. The Publisher will be happy to entertain applications from Schoolmasters for specimen copies of any...
Side 583 - If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means...
Side 574 - If a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram shall be double of the triangle.
Side 574 - Equal triangles, on equal bases, in the same straight line, and on the same side of it, are between the same parallels. Let...