Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J. Pryde. [With] Key1860 |
From inside the book
Results 1-5 of 11
Page 69
... harmonically . Let the line MN be cut into three segments in P and Q , such that MN PQ MPQN ; then will MN : NQ : MP : PQ . Since MN PQ = MPQN , therefore ( VI . 16 ) the sides of these rectangles are reciprocally pro- portional ...
... harmonically . Let the line MN be cut into three segments in P and Q , such that MN PQ MPQN ; then will MN : NQ : MP : PQ . Since MN PQ = MPQN , therefore ( VI . 16 ) the sides of these rectangles are reciprocally pro- portional ...
Page 70
... harmonically . Let AB be the given line to be divided harmonically . - CD ; join CB , Draw any line AE , and on it take any two segments AD and DC , and make CE : and draw DF parallel to CB ; join EF , and draw CG parallel to EF , and ...
... harmonically . Let AB be the given line to be divided harmonically . - CD ; join CB , Draw any line AE , and on it take any two segments AD and DC , and make CE : and draw DF parallel to CB ; join EF , and draw CG parallel to EF , and ...
Page 71
... harmonically . Or , more concisely thus : AB : BF = = = AC : CE AG : GF ( VI . 4 ) ; BF : GF ( V. 16 ) . = = AC : CD ( VI . 4 ) hence AB : BF = AG : GF , or AB : AG SCHOL . It is evident that the line AB could be cut har- monically in ...
... harmonically . Or , more concisely thus : AB : BF = = = AC : CE AG : GF ( VI . 4 ) ; BF : GF ( V. 16 ) . = = AC : CD ( VI . 4 ) hence AB : BF = AG : GF , or AB : AG SCHOL . It is evident that the line AB could be cut har- monically in ...
Page 73
... harmonically . Let ABC be a triangle , and D the bisection of the base CA , then if a line be drawn through D , cutting BC in G , and AB produced in E , and the line BO drawn through the vertex parallel to the base in O ; it is required ...
... harmonically . Let ABC be a triangle , and D the bisection of the base CA , then if a line be drawn through D , cutting BC in G , and AB produced in E , and the line BO drawn through the vertex parallel to the base in O ; it is required ...
Page 74
... harmonically . In either of the figures , since BO is parallel to AD , the angle BOE = ADE , and the angle at E is common to the two triangles E B A 0 B D 4 E EOB and EDA , they are therefore similar ; also the triangles GOB and GDC are ...
... harmonically . In either of the figures , since BO is parallel to AD , the angle BOE = ADE , and the angle at E is common to the two triangles E B A 0 B D 4 E EOB and EDA , they are therefore similar ; also the triangles GOB and GDC are ...
Other editions - View all
Euclid's Elements of Plane Geometry [Book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde No preview available - 2023 |
Euclid's Elements of Plane Geometry [book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde No preview available - 2018 |
Common terms and phrases
AB² AC² AD² altitude angle ACB BC² BD² bisects the angle centre chord circumference consequently construction cut harmonically describe a circle diagonals diameter dicular draw equal angles equiangular equilateral triangle EXERCISE exterior angle figure find a point find the locus given angle given circle given line given point greater half hence hypotenuse intersection isosceles triangle Let ABC line joining lines be drawn lines drawn opposite sides Pages parallelogram perpen perpendicular Price produced quadrilateral radius rectangle rectangle contained required locus required point required to prove required triangle right angles right-angled triangle Scholium segments semiperimeter side AC square straight line tangent touch triangle ABC Trig vertex vertical angle whence wherefore Wood-cuts
Popular passages
Page 72 - ABC be a triangle, and DE a straight line drawn parallel to the base BC ; then will AD : DB : : AE : EC.
Page 19 - The line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of the third side.
Page 55 - 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.
Page 26 - Prove that three times the sum of the squares on the sides of a triangle is equal to four times the sum of the squares on the lines drawn from the vertices to the middle points of the opposite sides.
Page 73 - If three quantities are in continued proportion, the first is to the third as the square of the first is to the square of the second. Let a : b = b : c. Then, 2=*. b с Therefore, ^xb. = ^x± b с bb Or °r
Page 58 - EH parallel to AB or DC, and through F draw FK parallel to AD or BC ; therefore each of the figures, AK, KB, AH, HD, AG, GC, BG...
Page 29 - The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the diagonals...
Page 24 - If from the middle point of one of the sides of a right-angled triangle, a perpendicular be drawn to the hypotenuse, the difference of the squares on the segments into which it is divided, is equal to the square on the other side.
Page 2 - Of all triangles having the same vertical angle, and whose bases pass through a given point, the least is that whose base is bisected in the given point.