A Text-book of Geometrical DeductionsLongmans, Green and Company, 1891 - Geometry |
From inside the book
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Page 4
... joining the mid - points of the sides of an equilateral triangle is itself equilateral . E Let D , E , F be the mid - points of the sides BC , CA , AB , of the equi- lateral A ABC ; it is required to prove that A DEF is equilateral . Β ...
... joining the mid - points of the sides of an equilateral triangle is itself equilateral . E Let D , E , F be the mid - points of the sides BC , CA , AB , of the equi- lateral A ABC ; it is required to prove that A DEF is equilateral . Β ...
Page 5
... Join Al . Show , as in Ex . 5 , that BI - CI . Use Euc . I. 4 to prove △ ABI = △ ACI ; and hence show that Al is ... joining the angular points of an isosceles triangle to the mid - points of the opposite sides meet in a point . Use ...
... Join Al . Show , as in Ex . 5 , that BI - CI . Use Euc . I. 4 to prove △ ABI = △ ACI ; and hence show that Al is ... joining the angular points of an isosceles triangle to the mid - points of the opposite sides meet in a point . Use ...
Page 6
... joins the vertex of an isosceles triangle to the mid - point of the base , bisects the vertical angle and is perpendicular to the base . Let ABC be an isosceles triangle having AB AC , and let BD = CD ; it is required to prove A ( 1 ) ...
... joins the vertex of an isosceles triangle to the mid - point of the base , bisects the vertical angle and is perpendicular to the base . Let ABC be an isosceles triangle having AB AC , and let BD = CD ; it is required to prove A ( 1 ) ...
Page 8
... Join OD . As in § 1 , Ex . 5 , show that AO BO = CO . Use Euc . I. 8 to prove △ BOD = A COD , and hence show ODBC ... joins the vertices of two isosceles triangles on the same base bisects both vertical angles . 9. ABCD is a ...
... Join OD . As in § 1 , Ex . 5 , show that AO BO = CO . Use Euc . I. 8 to prove △ BOD = A COD , and hence show ODBC ... joins the vertices of two isosceles triangles on the same base bisects both vertical angles . 9. ABCD is a ...
Page 12
... Join EC . AD = ED , DB = DC , [ Construction . In As ADB , EDC LADB = LEDC ; ... AADBA EDC ; ... AB = CE . But AC + CE > AE ; [ Hypothesis . [ Euc . I. 15 . [ Euc . I. 4 . [ Euc . I. 20 . DEFINITION . ... AC + AB > 2AD , or AD ( AB + AC ) ...
... Join EC . AD = ED , DB = DC , [ Construction . In As ADB , EDC LADB = LEDC ; ... AADBA EDC ; ... AB = CE . But AC + CE > AE ; [ Hypothesis . [ Euc . I. 15 . [ Euc . I. 4 . [ Euc . I. 20 . DEFINITION . ... AC + AB > 2AD , or AD ( AB + AC ) ...
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A Text-Book of Geometrical Deductions: Book I. Corresponding to ..., Book 1 James Blaikie,W. Thomson No preview available - 2017 |
Common terms and phrases
26 to show 38 to show ABCD altitude angle equal angular points apply Euc bisect bisectors Bookwork centre Compare Ex Construct a right-angled Construct a triangle Construct an isosceles convex polygon diagonals Draw a straight drawn parallel equal angles equilateral triangle EUCLID exterior angles Find a point Find the locus fixed point given line given point given square given straight line given the base given triangle hypotenuse isosceles triangle joining the mid-points LADC Let ABC line which joins lines be drawn median meet BC method of Ex mid-point of BC obtuse opposite angles opposite sides parallel straight lines parallelogram perimeter point in BC previous Ex quadrilateral quadrilateral ABCD rectangle required to prove respectively equal rhombus right angles right-angled triangle satisfies the condition Standard Theorem straight line drawn trapezium triangle required Trisect vertex vertical angle
Popular passages
Page 81 - In every triangle, the square on the side subtending an acute angle is less than the sum of the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle.
Page 27 - If two triangles have two sides of the one equal to two sides of the...
Page 135 - PROB. from a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line : it is required to draw from the point A a straight line equal to BC.
Page 136 - If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.
Page 138 - If the square described on one side of a triangle be equal to the sum of the squares described on the other two sides, the angle contained by these two sides is a right angle.
Page 81 - In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side...
Page 137 - THE straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.
Page 50 - A line which joins the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
Page 137 - ... upon the same side together equal to two right angles; the two straight lines shall be parallel to one another.
Page 135 - The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another.