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ABC is equal ABCD angle ABC angle BAC arch base bisect BOOK called centre chord circle circumference coincide common cone Consequently cylinder demonstrations described diagonals diameter difference distance divided draw drawn equal angles equal bases equiangular exterior angle extremities figure follows fore geometry given straight line gles greater half Hence interior intersect join less Let ABC magnitude manner mean meet opposite opposite angles opposite side paral parallel parallelogram pass perp perpendicular plane polygon prism PROBLEM produced Prop proportional PROPOSITION proved pyramid Q. E. D. Cor quantities radius ratio rectangle contained remaining respects right angles segments side AC sides similar solid square stand surface tangent THEOREM third triangle ABC unequal wherefore whole
Página ii - IDE, of the said District, hath deposited in this office, the title of a book, the right whereof he claims as proprietor, in the words following, to wit : " Inductive Grammar, designed for beginners. By an Instructer." In conformity to the act of the Congress of the United States...
Página 80 - If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.
Página 42 - 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...
Página 30 - To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line : it is required to divide it intotwo equal parts.
Página 20 - LET it be granted that a straight line may be drawn from any one point to any other point.
Página 38 - Problem. At a given point in a given straight line to make a rectilineal angle equal to a given rectilineal angle.
Página 113 - Wherefore also the angle BAD is equal to the angle CAD : Therefore the angle BAC is cut into two equal angles by the straight line AD.
Página 24 - DE ; the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide...