Elements of Geometry, After Legendre, with a Selection of Geometrical Exercises, and Hints for the Solution of the Same |
Αναζήτηση στο βιβλίο
Αποτελέσματα 1 - 5 από τα 84.
Σελίδα 30
... . THEOREM . In an isosceles triangle , the angles opposite to the equal sides are equal . Let the side AB = AC ; then shall the angle C be equal to B. Join the vertex A , and D , the middle point 30 ELEMENTS OF GEOMETRY .
... . THEOREM . In an isosceles triangle , the angles opposite to the equal sides are equal . Let the side AB = AC ; then shall the angle C be equal to B. Join the vertex A , and D , the middle point 30 ELEMENTS OF GEOMETRY .
Σελίδα 31
... join DC . The angle DBC is ( by hypothesis ) equal to ACB ; and the two sides DB , BC are equal to the two sides AC , CB ; therefore , the triangle DBC ( Prop . VII . ) must be equal to the triangle ACB . But the part cannot be equal to ...
... join DC . The angle DBC is ( by hypothesis ) equal to ACB ; and the two sides DB , BC are equal to the two sides AC , CB ; therefore , the triangle DBC ( Prop . VII . ) must be equal to the triangle ACB . But the part cannot be equal to ...
Σελίδα 32
... join DC . C B F The triangle CBD is equal to ABC : for the angles CBD and CBA are right angles , the side CB is common , and the side BD = AB ; therefore , the triangles are equal ( Prop . VII . ) , and hence , the angle BCD = BCA ; but ...
... join DC . C B F The triangle CBD is equal to ABC : for the angles CBD and CBA are right angles , the side CB is common , and the side BD = AB ; therefore , the triangles are equal ( Prop . VII . ) , and hence , the angle BCD = BCA ; but ...
Σελίδα 34
... Join IA , IB ; one of these lines will cut the perpendicular in D ; from D draw DB ; we shall have DB = DA . But the straight line IB is less than ID + DB , and ID + DB = ID + DA = IA . Therefore , IBIA . Therefore , every point out of ...
... Join IA , IB ; one of these lines will cut the perpendicular in D ; from D draw DB ; we shall have DB = DA . But the straight line IB is less than ID + DB , and ID + DB = ID + DA = IA . Therefore , IBIA . Therefore , every point out of ...
Σελίδα 36
... join MA ; the two right angled triangles MAF and MAE have the hypothenuse MA in common , and ME = MF by hypothesis ; hence they are equal ( Prop . XIX . ) . Therefore , the angle MAE MAF , and MA divides the angle BAC in half , hence ...
... join MA ; the two right angled triangles MAF and MAE have the hypothenuse MA in common , and ME = MF by hypothesis ; hence they are equal ( Prop . XIX . ) . Therefore , the angle MAE MAF , and MA divides the angle BAC in half , hence ...
Άλλες εκδόσεις - Προβολή όλων
Elements of Geometry, After Legendre: With a Selection of Geometrical ... Charles Scott Venable Προβολή αποσπασμάτων - 1890 |
Elements of Geometry, After Legendre: With a Selection of Geometrical ... Adrien Marie Legendre,Charles Scott Venable Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2014 |
Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD altitude angle ACB angles equal apothem base bisect bisectrix Book centre chord circ circumference circumscribed circle common cone Construct convex surface cylinder decagon described diagonals diameter dicular diedral distance divided draw drawn edges equally distant equilateral triangle equivalent Exercises exterior angles Find the locus frustum Geometric given circle given line given point given straight line greater Hence homologous homologous sides hypothenuse isosceles join let fall line parallel loci measure median meet middle point number of sides opposite parallelogram parallelopipedon perimeter perpen perpendicular plane angles plane MN point of contact point of intersection polyedron prism problem Prop proportional PROPOSITION pyramid quadrilateral radii ratio rectangle regular polygon right angled triangle S-ABC SCHOLIUM segment similar solid angle sphere spherical triangle square tangent tetraedron THEOREM triangle ABC triangular prism triedral vertex vertices volume
Δημοφιλή αποσπάσματα
Σελίδα 255 - THE sphere is a solid terminated by a curve surface, all the points of which are equally distant from a point within, called the centre.
Σελίδα 49 - The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it 46 INTERCEPTS BY PARALLEL LINES.
Σελίδα 20 - Parallel straight lines are such as are in the same plane, and which, being produced ever so far both ways, do not meet.
Σελίδα 54 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Σελίδα 45 - 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.
Σελίδα 67 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.
Σελίδα 107 - The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides.
Σελίδα 300 - The volume of a frustum of a cone is equivalent to the sum of the volumes of three cones whose common altitude is the altitude of the frustum and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum.
Σελίδα 276 - Scholium. The spherical ungula, bounded by the planes AMB, ANB, is to the whole solid sphere, as the angle A is to four right angles. For, the lunes being equal, the spherical ungulas will also be equal ; hence two spherical ungulas are to each other, as the angles formed by the planes which bound them. PROPOSITION XVIII.
Σελίδα 9 - Ratio is the relation which one quantity bears to another in respect of magnitude, the comparison being made by considering what multiple, part, or parts, one is of the other.