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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Resultat 1-5 av 77
Side 9
... Hypoth . ) , and the angle A is common , therefore the angle ACF is equal to ABG ; and the angle AFC to AGB ; and the side FC is equal to the side GB ( by Prop . 4 ) . Therefore in the triangles BFC and CGB , the angle BFC is equal to ...
... Hypoth . ) , and the angle A is common , therefore the angle ACF is equal to ABG ; and the angle AFC to AGB ; and the side FC is equal to the side GB ( by Prop . 4 ) . Therefore in the triangles BFC and CGB , the angle BFC is equal to ...
Side 10
... Hypoth . ) , the triangles themselves , DBC and ACB , will be equal ( by Prop . 4 ) —a part to the whole , which is absurd : therefore neither of the sides BA nor AC is greater than the other , therefore they are equal . Hence every ...
... Hypoth . ) , the triangles themselves , DBC and ACB , will be equal ( by Prop . 4 ) —a part to the whole , which is absurd : therefore neither of the sides BA nor AC is greater than the other , therefore they are equal . Hence every ...
Side 11
... Hypoth . ) , the angles ACD and ADC are equal ( by Prop . 5 ) ; but ACD is greater than BCD A D B ( Ax . 9 ) , therefore ADC is greater than BCD ; but BDC is greater than ADC ( by Ax . 9 ) , and therefore BDC is greater than BCD : but ...
... Hypoth . ) , the angles ACD and ADC are equal ( by Prop . 5 ) ; but ACD is greater than BCD A D B ( Ax . 9 ) , therefore ADC is greater than BCD ; but BDC is greater than ADC ( by Ax . 9 ) , and therefore BDC is greater than BCD : but ...
Side 15
... Hypoth . ) ; therefore CBA and ABE are equal to CBA and ABD : take away the common part CBA , and ABE will be equal to ABD ; a part to the whole , which is absurd . Therefore BE is not in the direct line with CB ; and it can be ...
... Hypoth . ) ; therefore CBA and ABE are equal to CBA and ABD : take away the common part CBA , and ABE will be equal to ABD ; a part to the whole , which is absurd . Therefore BE is not in the direct line with CB ; and it can be ...
Side 21
Euclides T W Herbert. Because the right lines AG and AC are equal ( by Hypoth . and Constr . ) the angles ACG and AGC will be equal ( by Prop . 5 ) : but BGC is greater than AGC , and therefore greater than ACG , and also greater than ...
Euclides T W Herbert. Because the right lines AG and AC are equal ( by Hypoth . and Constr . ) the angles ACG and AGC will be equal ( by Prop . 5 ) : but BGC is greater than AGC , and therefore greater than ACG , and also greater than ...
Vanlige uttrykk og setninger
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
Populære avsnitt
Side 20 - If two triangles have two sides of the one equal to two sides of the...
Side 30 - DE : but equal triangles on the same base and on the same side of it, are between the same parallels ; (i.
Side 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.
Side 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.
Side 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.
Side 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.
Side 129 - In any proportion, the product of the means is equal to the product of the extremes.
Side 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.
Side 215 - ... are to one another in the duplicate ratio of their homologous sides.
Side 160 - PROPOSITION XV. PROBLEM. To inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle.