Plane and Solid GeometryLongmans, Green and Company, 1898 - 210 σελίδες |
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Σελίδα 9
... likewise DCB is a right angle , or ( by 28 ) / ECB = / DCB , A B C which is impossible , as DCB is a part of ECB . EXERCISE . If in Cor . 2 the angles EAF , CAD , and BAC are equal , and each twice as large as the angle DAE , what will ...
... likewise DCB is a right angle , or ( by 28 ) / ECB = / DCB , A B C which is impossible , as DCB is a part of ECB . EXERCISE . If in Cor . 2 the angles EAF , CAD , and BAC are equal , and each twice as large as the angle DAE , what will ...
Σελίδα 35
... Likewise , since LBAC LACE , = CAE = ZACB . and By addition , or ZBAEZ BCE , ZA = ZC . Q.E.D. 107. COR . 1. A diagonal of a parallelogram divides it into two equal triangles . 108. COR . 2. Parallel lines included between parallels are ...
... Likewise , since LBAC LACE , = CAE = ZACB . and By addition , or ZBAEZ BCE , ZA = ZC . Q.E.D. 107. COR . 1. A diagonal of a parallelogram divides it into two equal triangles . 108. COR . 2. Parallel lines included between parallels are ...
Σελίδα 37
... Likewise , FL = BC , and GS : = CD . But AB = BC = CD by hypothesis , then ( by 28 ) EK = FL = GS . M 0 A E B F K C G L D H S N In the triangles EKF , FLG , and GSH the angles KEF , LFG , and SGH are equal ( by 62 ) ; also the angles ...
... Likewise , FL = BC , and GS : = CD . But AB = BC = CD by hypothesis , then ( by 28 ) EK = FL = GS . M 0 A E B F K C G L D H S N In the triangles EKF , FLG , and GSH the angles KEF , LFG , and SGH are equal ( by 62 ) ; also the angles ...
Σελίδα 38
... Likewise ( by 111 ) , BC — DE = DE Q.E.D. AADE - 0 = DE . 30 or Transpose DE , then BC = 2 DE , DEBC . Q.E.D. PROPOSITION XXIV . THEOREM . 113. The line drawn parallel to the bases through the middle point of one of the non - parallel ...
... Likewise ( by 111 ) , BC — DE = DE Q.E.D. AADE - 0 = DE . 30 or Transpose DE , then BC = 2 DE , DEBC . Q.E.D. PROPOSITION XXIV . THEOREM . 113. The line drawn parallel to the bases through the middle point of one of the non - parallel ...
Σελίδα 40
... likewise BO BR . = = Q.E.D. By taking CP and BR as medial lines intersecting at O , and joining S , the middle point of CO , with N , and with R , and drawing RP and NP , it can be shown in the same manner that OC = = CP , and that O is ...
... likewise BO BR . = = Q.E.D. By taking CP and BR as medial lines intersecting at O , and joining S , the middle point of CO , with N , and with R , and drawing RP and NP , it can be shown in the same manner that OC = = CP , and that O is ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD AC² acute angle AD² adjacent adjacent angles altitude angle formed angles are equal apothem arc BC base and altitude bisect bisector called centre chord circumference circumscribed cone cylinder diagonals diameter diedral angles distance divided draw drawn ECDH equally distant equilateral equivalent EXERCISES faces four right angles frustum given point given straight line hence homologous homologous sides hypotenuse inscribed polygon interior angles intersection isosceles triangle join lateral area lateral edges Let ABC lune mean proportional measured by one-half middle point number of sides parallelogram parallelopiped perimeter perpendicular polyedral angle polyedron PROPOSITION XI prove pyramid Q.E.D. PROPOSITION quadrilateral radii radius ratio rectangle rectangular parallelopiped regular polygon right triangle SCHOLIUM segments semiperimeter sphere spherical angle spherical polygon spherical triangle surface tangent THEOREM triangle ABC triangles are equal triangular triangular prism V-ABC vertex vertical angle
Δημοφιλή αποσπάσματα
Σελίδα 46 - PERIPHERY of a circle is its entire bounding line ; or it is a curved line, all points of which are equally distant from a point within called the centre.
Σελίδα 105 - ... any two parallelograms are to each other as the products of their bases by their altitudes. PROPOSITION V. THEOREM. 403. The area of a triangle is equal to half the product of its base by its altitude.
Σελίδα 82 - If any number of quantities are proportional, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let a : b = c : d = e :f Now ab = ab (1) and by Theorem I.
Σελίδα 192 - A sphere is a solid bounded by a surface all points of which are equally distant from a point within called the centre.
Σελίδα 108 - Two triangles having 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.
Σελίδα 146 - A STRAIGHT line is perpendicular to a plane, when it is perpendicular to every straight line which it meets in that plane.
Σελίδα 30 - In an isosceles triangle, the angles opposite the equal sides are equal.
Σελίδα 80 - In any proportion the terms are in proportion by Composition ; that is, the sum of the first two terms is to the first term as the sum of the last two terms is to the third term.
Σελίδα 79 - If the product of two quantities is equal to the product of two others, one pair may be made the extremes, and the other pair the means, of a proportion. Let ad = ос.
Σελίδα 148 - Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of the perpendicular ; and of two unequal oblique lines the greater meets the plane at the greater distance from the foot of the perpendicular.