Euclid's Elements of geometry, the first three books (the fourth, fifth, and sixth books) tr. from the Lat. To which is added, A compendium of algebra (A compendium of trigonometry).1846 |
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Página 6
... than two right angles ; these two right lines , being produced , will meet one another on that side at which the angles are less than two right angles . PROPOSITION I. PROBLEM . On a given finite right line 6 FIRST BOOK .
... than two right angles ; these two right lines , being produced , will meet one another on that side at which the angles are less than two right angles . PROPOSITION I. PROBLEM . On a given finite right line 6 FIRST BOOK .
Página 7
... PROBLEM . From a given point ( A ) to draw a right line equal to a given finite right line ( BC ) . From the given point A draw a right line AB to either extremity B of the given right line ( by Post . 1 ) ; upon AB construct an ...
... PROBLEM . From a given point ( A ) to draw a right line equal to a given finite right line ( BC ) . From the given point A draw a right line AB to either extremity B of the given right line ( by Post . 1 ) ; upon AB construct an ...
Página 8
... PROBLEM . From the greater of two given right lines ( AB and CF ) to cut off a part equal to the less . From either extremity A of the greater right line , draw AD , equal to the less CF of the given lines ( by Prop . 2 ) . From the ...
... PROBLEM . From the greater of two given right lines ( AB and CF ) to cut off a part equal to the less . From either extremity A of the greater right line , draw AD , equal to the less CF of the given lines ( by Prop . 2 ) . From the ...
Página 12
... PROBLEM . To cut a given rectilineal angle ( BAC ) into two equal parts . From A take the equals AD and AE ( by Prop . 3 , ) draw DE , and upon it con- struct an equilateral triangle DFE ( by Prop . 1 ) . The right line joining the ...
... PROBLEM . To cut a given rectilineal angle ( BAC ) into two equal parts . From A take the equals AD and AE ( by Prop . 3 , ) draw DE , and upon it con- struct an equilateral triangle DFE ( by Prop . 1 ) . The right line joining the ...
Página 13
... PROBLEM . To draw a perpendicular to a given right line ( AB ) , from a point ( C ) in the given line . From the given point C make both CD and CE equal ( by Prop . 3 ) ; upon DE construct an equilateral triangle DFE ; draw FC , and it ...
... PROBLEM . To draw a perpendicular to a given right line ( AB ) , from a point ( C ) in the given line . From the given point C make both CD and CE equal ( by Prop . 3 ) ; upon DE construct an equilateral triangle DFE ; draw FC , and it ...
Términos y frases comunes
absurd AC and CB AC by Prop AC is equal angle ABC angle equal angles by Prop arch bisected centre circumference co-efficient Const construct contained oftener diameter divided divisor double equal angles equal by Constr equal by Hypoth equal by Prop equal right lines equal to AC equal to twice equi-multiples equi-submultiples equiangular equilateral external angle fore fraction given angle given circle given line given right line given triangle greater half a right inscribed less multiplied opposite parallel parallelogram perpendicular PROPOSITION quantities quotient ratio rectangle under AC remaining angles remaining side right angle right line AB right line AC SCHOL segment semicircle side AC similar similarly demonstrated squares of AC submultiple subtract THEOREM tiple touches the circle triangle BAC twice the rectangle twice the square whole
Pasajes populares
Página 20 - If two triangles have two sides of the one equal to two sides of the...
Página 30 - DE : but equal triangles on the same base and on the same side of it, are between the same parallels ; (i.
Página 209 - ... they have an angle of one equal to an angle of the other and the including sides are proportional; (c) their sides are respectively proportional.
Página 218 - 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.
Página 114 - To reduce fractions of different denominators to equivalent fractions having a common denominator. RULE.! Multiply each numerator into all the denominators except its own for a new numerator, and all the denominators together for a common denominator.
Página 90 - The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.
Página 129 - In any proportion, the product of the means is equal to the product of the extremes.
Página 163 - Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever be taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order.
Página 215 - ... are to one another in the duplicate ratio of their homologous sides.
Página 160 - PROPOSITION XV. PROBLEM. To inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle.