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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Page 10
Euclides T W Herbert. equals AB and AC from the equals AF and AG , the side BF is also equal to the side CG ; therefore the angle FBC is equal to the angle GCB ( by Prop . 4 ) ; but these are the angles below the base BC . And in the ...
Euclides T W Herbert. equals AB and AC from the equals AF and AG , the side BF is also equal to the side CG ; therefore the angle FBC is equal to the angle GCB ( by Prop . 4 ) ; but these are the angles below the base BC . And in the ...
Page 11
... equal ( by Hypoth . ) , therefore the angles BDC and BCD are equal ( by Prop . 6 ) ; but the angle BDC is greater than BCD , which is absurd . Therefore the triangles constructed on the same right line cannot have their conterminous sides ...
... equal ( by Hypoth . ) , therefore the angles BDC and BCD are equal ( by Prop . 6 ) ; but the angle BDC is greater than BCD , which is absurd . Therefore the triangles constructed on the same right line cannot have their conterminous sides ...
Page 12
... equal sides AB and EF , CB and DF may be conterminous , the vertex B must fall on F ( by Prop . 7 ) ; and the equal sides , AB and EF , CB and FD must agree , ( by Ax . 10 ) ; therefore , the angles B and F must coincide , and there ...
... equal sides AB and EF , CB and DF may be conterminous , the vertex B must fall on F ( by Prop . 7 ) ; and the equal sides , AB and EF , CB and FD must agree , ( by Ax . 10 ) ; therefore , the angles B and F must coincide , and there ...
Page 13
... equal ( by Constr . ) , but CD is common , and the angles ACD , BCD are equal ( by Constr . ) , therefore the bases AD and DB are equal ( by Prop . 4 ) , and therefore the given right line A B is bisected in D. PROPOSITION XI . PROBLEM ...
... equal ( by Constr . ) , but CD is common , and the angles ACD , BCD are equal ( by Constr . ) , therefore the bases AD and DB are equal ( by Prop . 4 ) , and therefore the given right line A B is bisected in D. PROPOSITION XI . PROBLEM ...
Page 15
... equal to two right angles ; these right lines ( CB and BD ) will form one straight line . For if not , if it be possible , let BE be in the same right line with CB , and the angles CBA E D and ABE will be equal to two right angles ( by Prop ...
... equal to two right angles ; these right lines ( CB and BD ) will form one straight line . For if not , if it be possible , let BE be in the same right line with CB , and the angles CBA E D and ABE will be equal to two right angles ( by Prop ...
Common terms and phrases
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
Popular passages
Page 20 - If two triangles have two sides of the one equal to two sides of the...
Page 30 - DE : but equal triangles on the same base and on the same side of it, are between the same parallels ; (i.
Page 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.
Page 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.
Page 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.
Page 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.
Page 129 - In any proportion, the product of the means is equal to the product of the extremes.
Page 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.
Page 215 - ... are to one another in the duplicate ratio of their homologous sides.
Page 160 - PROPOSITION XV. PROBLEM. To inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle.