Euclid's Elements of plane geometry [book 1-6] with explanatory appendix, and supplementary propositions, by W.D. Cooley1840 |
Inni boken
Resultat 1-5 av 16
Side 40
... twice as many right angles ( deducting four ) as the figure has sides . B с E For , if from any point F , taken within the figure , the straight lines FA , FB , FC , & c . be drawn to the extre- mities of the sides , the figure will be ...
... twice as many right angles ( deducting four ) as the figure has sides . B с E For , if from any point F , taken within the figure , the straight lines FA , FB , FC , & c . be drawn to the extre- mities of the sides , the figure will be ...
Side 51
... twice the rectangle contained by the parts . = C E D K G F B Upon AB describe the square ACDB ; draw the diago- nal AD ; from F draw FE parallel to AC , and through the point G , where it cuts the diagonal , draw HK parallel to AB . The ...
... twice the rectangle contained by the parts . = C E D K G F B Upon AB describe the square ACDB ; draw the diago- nal AD ; from F draw FE parallel to AC , and through the point G , where it cuts the diagonal , draw HK parallel to AB . The ...
Side 53
... twice the rectangle contained by the whole line and that part , together with the square of the other part . C E D Upon AB construct the square ACDB ; draw the diagonal AD ; from F draw FE parallel to AC , and cutting the diagonal in G ...
... twice the rectangle contained by the whole line and that part , together with the square of the other part . C E D Upon AB construct the square ACDB ; draw the diagonal AD ; from F draw FE parallel to AC , and cutting the diagonal in G ...
Side 55
... twice AC2 , since AC = CE . And also , because EGF = / ECB ( i . Prop . 29 ) , the former is a right angle , and EF2 = EG2 + GF2 , or twice GF2 , since EG = GF , or twice CD2 , since GF = CD ( i . Prop . 34 ) . Therefore , AF2 = 2 AC2 + ...
... twice AC2 , since AC = CE . And also , because EGF = / ECB ( i . Prop . 29 ) , the former is a right angle , and EF2 = EG2 + GF2 , or twice GF2 , since EG = GF , or twice CD2 , since GF = CD ( i . Prop . 34 ) . Therefore , AF2 = 2 AC2 + ...
Side 57
... twice the rectangle contained by either ( AB ) of those sides and the produced part of it ( BD ) intercepted between the perpendicular ( CD ) let fall on it from the opposite angle , and the vertex of the obtuse angle . Since AD is ...
... twice the rectangle contained by either ( AB ) of those sides and the produced part of it ( BD ) intercepted between the perpendicular ( CD ) let fall on it from the opposite angle , and the vertex of the obtuse angle . Since AD is ...
Vanlige uttrykk og setninger
ACDB adjacent angles angles equal antecedent Axioms base bisected centre chord circumference coincide consequently Const definition demonstrated describe diagonal diameter difference divided draw equal angles equal Prop equal sides equiangular equilateral triangle equimultiples Euclid Euclid's Elements external angle extremities fore fourth fractional Geometry given angle given circle given line given point given straight line given triangle greater hypotenuse inscribed internal intersect isosceles triangle less line drawn lines be drawn magnitudes manner meeting multiple opposite angles parallel parallelogram perpendicular point of contact PROB produced proportional Proposition quadrilateral figure rectangle contained rectilinear figure remaining angles respectively equal right angle segment semiperimeter sides AC sides equal square of half subtending taken tangent THEOR third triangles ABC unequal vertex whole line
Populære avsnitt
Side 126 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Side 155 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Side 83 - If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tangent are equal to the angles in the alternate segments.
Side 129 - ... figures are to one another in the duplicate ratio of their homologous sides.
Side 47 - DE : but equal triangles on the same base and on the same side of it, are between the same parallels ; (i.
Side 90 - BFE : (i. def. 10.) therefore, in the two triangles, EAF, EBF, there are two angles in the one equal to two angles in the other, each to each ; and the side EF, which is opposite to one of the equal angles in each, is common to both ; therefore the other sides are equal ; (i.
Side 117 - A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less.
Side 56 - If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.
Side 60 - Iff a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.
Side 78 - Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another. If it be possible. let the two similar segments of circles, viz. ACB' ADB be upon the same side of the same straight line AB, not coinciding with one another.