The first six books of the Elements of Euclid, and propositions i.-xxi. of book xi1885 |
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Página 35
... manner AC is greater than EC . Therefore the sum of BA , AC is greater than BČ . Exercises . 1. In any triangle , the difference between any two sides is less than the third . 2. If any point within a triangle be joined to its angular ...
... manner AC is greater than EC . Therefore the sum of BA , AC is greater than BČ . Exercises . 1. In any triangle , the difference between any two sides is less than the third . 2. If any point within a triangle be joined to its angular ...
Página 38
... manner GK is equal to C , and FG is equal to B ( const . ) Hence A B с F G HE the three sides of the triangle KFG are respectively equal to the three lines A , B , C. Questions for Examination . 1. What is the reason for stating in the ...
... manner GK is equal to C , and FG is equal to B ( const . ) Hence A B с F G HE the three sides of the triangle KFG are respectively equal to the three lines A , B , C. Questions for Examination . 1. What is the reason for stating in the ...
Página 48
... manner the angle GHF is equal to HKD [ XXIX . ] . Therefore the angle AGK is equal to the angle GKD ( Axiom 1. ) . Hence [ XXVII . ] AB is parallel to CD . Through PROP . XXXI . - PROBLEM . B H F K a given poini ( C ) to draw a right ...
... manner the angle GHF is equal to HKD [ XXIX . ] . Therefore the angle AGK is equal to the angle GKD ( Axiom 1. ) . Hence [ XXVII . ] AB is parallel to CD . Through PROP . XXXI . - PROBLEM . B H F K a given poini ( C ) to draw a right ...
Página 55
... manner , since the parallelograms HB , HF are on the same base EH , and between the same parallels EH , BG , they are equal . Hence BD and FH are each equal to BH . Therefore ( Axiom 1. ) BD is equal to FH . Exercise . - Prove this ...
... manner , since the parallelograms HB , HF are on the same base EH , and between the same parallels EH , BG , they are equal . Hence BD and FH are each equal to BH . Therefore ( Axiom 1. ) BD is equal to FH . Exercise . - Prove this ...
Página 56
... manner the triangle DBC is half the parallelogram DBCF , because the diagonal DC bisects it , and halves of equal things are equal ( Axiom VII . ) . Therefore the triangle ABC is equal to the triangle DBC . Exercises . 1. If two equal ...
... manner the triangle DBC is half the parallelogram DBCF , because the diagonal DC bisects it , and halves of equal things are equal ( Axiom VII . ) . Therefore the triangle ABC is equal to the triangle DBC . Exercises . 1. If two equal ...
Términos y frases comunes
ABCD AC is equal AD² adjacent angles altitude angle ABC angle ACB angle BAC angular points Axiom bisector bisects centre chord circles touch circumference circumscribed circle collinear concurrent lines const coplanar cyclic quadrilateral Dem.-Let diagonals diameter divided draw equal angles equal to AC equiangular equilateral triangle escribed circles Euclid Exercises exterior angle Geometry given circle given line given point greater Hence the angle hypotenuse inscribed isosceles less line AC line joining locus manner meet middle points multiple nine-points circle opposite sides parallel parallelogram parallelopiped perpendicular plane points of intersection prism PROP Proposition prove radii radius rectangle contained rectilineal figure regular polygon respectively equal right angles right line segments semicircle sides AC similar square on AC tangent theorem triangle ABC vertex vertical angle
Pasajes populares
Página 295 - Thus the proposition, that the sum of the three angles of a triangle is equal to two right angles, (Euc.
Página 182 - When of the equimultiples of four magnitudes (taken as in the fifth definition) the multiple of the first is greater than that of the second, but the multiple of the third is not greater than the multiple of the fourth ; then the first is said to have to the second a greater ratio than the third magnitude has to the fourth...
Página 9 - LET it be granted that a straight line may be drawn from any one point to any other point.
Página 102 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square on the other part.
Página 122 - The diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is always greater than one more remote; and the greater is nearer to the centre than the less. Let ABCD be a circle, of which...
Página 226 - If from any angle of a triangle, a straight line be drawn perpendicular to the base ; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.
Página 29 - Again ; the mathematical postulate, that " things which are equal to the same are equal to one another," is similar to the form of the syllogism in logic, which unites things agreeing in the middle term.
Página 63 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Página 126 - The diagonals of a quadrilateral intersect at right angles. Prove that the sum of the squares on one pair of opposite sides is equal to the sum of the squares on the other pair.
Página 194 - If there be any number of proportionals, as one antecedent is to its consequent, so is the sum of all the antecedents to the sum of all the consequents.