Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J. Pryde. [With] Key1860 |
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Side 1
... hence the triangles are equal in all respects ( I. 4 ) , therefore AD = BD , and the angle ADC = BDC ; but they are adjacent angles ; therefore each of them is a right angle , hence CD is perpendicular to AB ( I. Def . 11 ) . A D B ...
... hence the triangles are equal in all respects ( I. 4 ) , therefore AD = BD , and the angle ADC = BDC ; but they are adjacent angles ; therefore each of them is a right angle , hence CD is perpendicular to AB ( I. Def . 11 ) . A D B ...
Side 3
... hence PC is the least line that can be drawn from P to the line AB , and PD is less than PE , and PE is less than PA . Also , there can only be drawn two equal lines from P to AB , one on each side of the perpendicular PC ; for make CB ...
... hence PC is the least line that can be drawn from P to the line AB , and PD is less than PE , and PE is less than PA . Also , there can only be drawn two equal lines from P to AB , one on each side of the perpendicular PC ; for make CB ...
Side 4
... hence AB bisects PQ , and cuts it at right angles . EXERCISE VII . - THEOREM . D In the figure , Euclid I. 5 , if BG and CF meet in O , shew that OA bisects the angle BAC . Referring to the figure of ( I. 5 ) , in the triangles FBC and ...
... hence AB bisects PQ , and cuts it at right angles . EXERCISE VII . - THEOREM . D In the figure , Euclid I. 5 , if BG and CF meet in O , shew that OA bisects the angle BAC . Referring to the figure of ( I. 5 ) , in the triangles FBC and ...
Side 7
... hence AD , DB , and DC are all equal . EXERCISE XV . - THEOREM . 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 that point . Let ABC and DBE be ...
... hence AD , DB , and DC are all equal . EXERCISE XV . - THEOREM . 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 that point . Let ABC and DBE be ...
Side 8
... hence DGDF ; in the same way it can be demon- strated from the triangles DEB and DFB that_DE = DF , and since DG and DE are each equal to DF , DG = DE . Again , in the two triangles DAG and DAE , right angled at G and E , the side DG ...
... hence DGDF ; in the same way it can be demon- strated from the triangles DEB and DFB that_DE = DF , and since DG and DE are each equal to DF , DG = DE . Again , in the two triangles DAG and DAE , right angled at G and E , the side DG ...
Andre utgaver - Vis alle
Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J ... Euclides Uten tilgangsbegrensning - 1860 |
Euclid's Elements of Plane Geometry [Book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Ingen forhåndsvisning tilgjengelig - 2023 |
Euclid's Elements of Plane Geometry [book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Ingen forhåndsvisning tilgjengelig - 2018 |
Vanlige uttrykk og setninger
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
Populære avsnitt
Side 72 - ABC be a triangle, and DE a straight line drawn parallel to the base BC ; then will AD : DB : : AE : EC.
Side 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.
Side 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.
Side 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.
Side 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
Side 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...
Side 29 - The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the diagonals...
Side 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.
Side 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.