The Element of GeometryW.E. Dean, 1836 - 114 Seiten |
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Ergebnisse 6-10 von 22
Seite 36
... reason HF also is a square , and it is upon the side HG which is equal to AC ; therefore HF , CK are the squares of AC , CB : and because the com- plement AG is equal ( 19. 2. ) to the complement GE , and that AG is the rectangle ...
... reason HF also is a square , and it is upon the side HG which is equal to AC ; therefore HF , CK are the squares of AC , CB : and because the com- plement AG is equal ( 19. 2. ) to the complement GE , and that AG is the rectangle ...
Seite 39
... reason , each of the angles CEB , EBC is half a right angle ; therefore AEB is a right angle ; and because EBC is half a right angle , DBG is also ( 3. 1. ) half a right angle , for they are vertically opposite ; but BDG is a right ...
... reason , each of the angles CEB , EBC is half a right angle ; therefore AEB is a right angle ; and because EBC is half a right angle , DBG is also ( 3. 1. ) half a right angle , for they are vertically opposite ; but BDG is a right ...
Seite 48
... reason CD is double of CG , and AB is equal to CD , therefore AF is equal to CG ; and because AE is equal to EC , the square of AE is equal to the square of EC ; but the squares of AF , FE are equal ( 17. 2. ) to the square of AE ...
... reason CD is double of CG , and AB is equal to CD , therefore AF is equal to CG ; and because AE is equal to EC , the square of AE is equal to the square of EC ; but the squares of AF , FE are equal ( 17. 2. ) to the square of AE ...
Seite 52
... reason , the angle FEC is double of the angle EAC ; therefore the whole angle BEC is double of the whole angle BAC . A B F E A D E G B Again , Let BDC be inflectedt o the circumference , so that E the cen- tre of the circle be without ...
... reason , the angle FEC is double of the angle EAC ; therefore the whole angle BEC is double of the whole angle BAC . A B F E A D E G B Again , Let BDC be inflectedt o the circumference , so that E the cen- tre of the circle be without ...
Seite 53
... reason , be- cause CBED is greater than a semicircle , the angles CAD , CED are equal : there- fore the whole angle BAD is equal to the whole angle BED . Wherefore the angles in the same segment , & c . Q. E. D. COR . Segments of ...
... reason , be- cause CBED is greater than a semicircle , the angles CAD , CED are equal : there- fore the whole angle BAD is equal to the whole angle BED . Wherefore the angles in the same segment , & c . Q. E. D. COR . Segments of ...
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The Element of Geometry (Classic Reprint) Formerly Chairman Department of Immunology John Playfair,John Playfair Keine Leseprobe verfügbar - 2017 |
Häufige Begriffe und Wortgruppen
AB is equal AC is equal adjacent angles angle ABC angle ACB angle BAC angle DEF arc AC base BC bisect called centre circle ABCD circle EFGH coincide described divided drawn duplicate ratio equal 17 equal angles equal circles equal to CD equiangular FGHKL fore four right angles given straight line gnomon greater homologous sides join less Let ABC Let the straight opposite angles parallel parallelogram perpendicular polygon PROB produced Q. E. D. PROP radius rectangle BC rectangle contained rectilineal figure remaining angle right angled triangle secant segment side AC similar sine square of AC straight line AB straight line AC THEOR touches the circle triangle ABC twice the rectangle wherefore whole angle
Beliebte Passagen
Seite 25 - If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Seite 13 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw a straight line E iR.
Seite 4 - To draw a straight line at right angles to a given straight line, from a given point in the same.
Seite 29 - In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side...
Seite 90 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds ; and these into thirds, &c.
Seite 87 - ... magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes.
Seite 1 - If two triangles have two sides of the one equal to two sides of the...
Seite 13 - If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles.
Seite 64 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Seite 19 - If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...