The elements of plane geometry, from the Sansk. text of Ayra Bhatta, ed. by Jasoda Nandan Sircar1878 |
Inni boken
Resultat 1-5 av 23
Side i
... proved by the next following . The latter is proved by the next following it and so on . This inverse arrangement has rendered the subject unnecessarily abstruse and definitely points out to the learner the necessity of attending ...
... proved by the next following . The latter is proved by the next following it and so on . This inverse arrangement has rendered the subject unnecessarily abstruse and definitely points out to the learner the necessity of attending ...
Side 7
... prove that the diameter bisects the circle . This is simply the converse of the definition of a circle . The two corollaries are also to be proved by the same definition . For if ADC be applied to ABC , they coincide ( 7 )
... prove that the diameter bisects the circle . This is simply the converse of the definition of a circle . The two corollaries are also to be proved by the same definition . For if ADC be applied to ABC , they coincide ( 7 )
Side 11
... proved to coin- cide with the point E. Therefore the base BC shall coincide with the base EF . For , the point B coinciditng with the point E , and the point C with the point F , if the base BC does not coincide with the base EF , the ...
... proved to coin- cide with the point E. Therefore the base BC shall coincide with the base EF . For , the point B coinciditng with the point E , and the point C with the point F , if the base BC does not coincide with the base EF , the ...
Side 12
... proved , that no part of AB can be equal to AC . For , if possible , let BD a part of AB be equal to AC . Join DC . Because in the two triangles DBC , ACB , the side DB is equal to the side AC ( hyp ) , and BC is common to both , the ...
... proved , that no part of AB can be equal to AC . For , if possible , let BD a part of AB be equal to AC . Join DC . Because in the two triangles DBC , ACB , the side DB is equal to the side AC ( hyp ) , and BC is common to both , the ...
Side 13
... proved , that the angle HAB is equal to the angle HBA ( P. 4 ) . Therefore the whole angle GAH is equal to the whole angle GBH ( Ax . 2 ) . Because in the two triangles AGH and BGH the sides AG , AH and BG , BH are equal and the ...
... proved , that the angle HAB is equal to the angle HBA ( P. 4 ) . Therefore the whole angle GAH is equal to the whole angle GBH ( Ax . 2 ) . Because in the two triangles AGH and BGH the sides AG , AH and BG , BH are equal and the ...
Vanlige uttrykk og setninger
AB is equal AC is equal angle ABC angle ADB angle BAC angle DEF angle EDF angles are equal arc BC base BC base CD bisected centre circle ABCD circumference coincide conversely diameter draw duplicate ratio equal angles equal Ax equal to AC equiangular equimultiples Euclid exterior angle figure described fore given straight lines greater Hindoo Geometry homologous isosceles triangle Join Let ABC mean proportionals multiple opposite angles parallel parallelogram AC perpendicular point F polygon produced Q. E. D. Cor Q.E.D. PROP rectangle contained rectilineal figure regular polygon remaining angle right angles sector BGC sector EHF segment semicircle side BC square of BC straight line &c THEOREM third angle touches the circle triangle ABC Wherefore whole angle
Populære avsnitt
Side 10 - If two triangles have two sides of the one equal to two sides of the...
Side 39 - 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 5 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
Side 40 - ... figure, together with four right angles, are equal to twice as many right angles as the figure has be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles.
Side 79 - THE rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides.* Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD ; the.
Side 74 - Any two sides of a triangle are together greater than the third side.
Side 47 - In every triangle, the square on the side subtending either of the acute angles, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the...
Side 53 - ... figures are to one another in the duplicate ratio of their homologous sides.
Side 62 - ... in a segment less than a semicircle, is greater than a right angle...
Side 59 - The angles in the same segment of a circle are equal to one another.