A Text-book of Geometrical DeductionsLongmans, Green and Company, 1891 |
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Resultat 1-5 av 15
Side
... Perimeter , 14 Bisector ,. 19 Distance from a Line , 20 Convex Polygon , 42 Centre of Parallelogram , 56 Orthocentre , Trapezium , Locus , Inscribed Figures , 60 62 92 110 · : || ||| SYMBOLS . signifies therefore . 99 is equal to . is ...
... Perimeter , 14 Bisector ,. 19 Distance from a Line , 20 Convex Polygon , 42 Centre of Parallelogram , 56 Orthocentre , Trapezium , Locus , Inscribed Figures , 60 62 92 110 · : || ||| SYMBOLS . signifies therefore . 99 is equal to . is ...
Side 14
... Perimeter . B Thus BC + CA + AB is the perimeter of the A ABC . 8. The sum of the medians of a triangle is less than the peri- meter of the triangle . A C Use Ex . 1 for the three medians in turn , add the results , and divide by 2 ...
... Perimeter . B Thus BC + CA + AB is the perimeter of the A ABC . 8. The sum of the medians of a triangle is less than the peri- meter of the triangle . A C Use Ex . 1 for the three medians in turn , add the results , and divide by 2 ...
Side 15
... perimeter of the triangle , and ( 2 ) less than the perimeter . Let P be any point within △ ABC ; it is required to prove that ( 1 ) PA + PB + PC > ( BC + CA + AB ) , ( 2 ) PA + PB + PC < BC + CA + AB . B ( 1 ) In △ PBC , PB + PC > BC ...
... perimeter of the triangle , and ( 2 ) less than the perimeter . Let P be any point within △ ABC ; it is required to prove that ( 1 ) PA + PB + PC > ( BC + CA + AB ) , ( 2 ) PA + PB + PC < BC + CA + AB . B ( 1 ) In △ PBC , PB + PC > BC ...
Side 16
James Andrew Blaikie, William Thomson. 2. The perimeter of a quadrilateral is ( 1 ) greater than the sum of the diagonals , and ( 2 ) less than twice the sum of the diagonals . B A A For ( 1 ) use Euc . I. 20 to prove that C AB + BC > AC ...
James Andrew Blaikie, William Thomson. 2. The perimeter of a quadrilateral is ( 1 ) greater than the sum of the diagonals , and ( 2 ) less than twice the sum of the diagonals . B A A For ( 1 ) use Euc . I. 20 to prove that C AB + BC > AC ...
Side 17
... perimeter . A For ( 1 ) use Euc . I. 20 . PA + PB > AB , etc. For ( 2 ) show , as in Ex . 4 , that BC + CD + DA > PA + PB , etc. 6. If within a quadrilateral ABCD , P be a point which is not the point of intersection of the diagonals ...
... perimeter . A For ( 1 ) use Euc . I. 20 . PA + PB > AB , etc. For ( 2 ) show , as in Ex . 4 , that BC + CD + DA > PA + PB , etc. 6. If within a quadrilateral ABCD , P be a point which is not the point of intersection of the diagonals ...
Andre utgaver - Vis alle
A Text-book of Geometrical Deductions: Corresponding to Euclid, book ..., Bok 2 James Blaikie Uten tilgangsbegrensning - 1892 |
A Text-book of Geometrical Deductions James Andrew Blaikie,William Thomson Uten tilgangsbegrensning - 1891 |
A Text-Book of Geometrical Deductions: Book I. Corresponding to ..., Bok 1 James Blaikie,W. Thomson Ingen forhåndsvisning tilgjengelig - 2017 |
Vanlige uttrykk og setninger
26 to show 38 to show ABCD altitude angle equal angular points apply Euc bisectors Bookwork centre Construct a right-angled Construct a triangle Construct an isosceles contained angle convex polygon diagonals Draw a straight drawn parallel equal angles equilateral triangle EUCLID exterior angles Find a point find the locus fixed point given angle given line given point given square given straight line given triangle gonals 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 opposite angles opposite sides parallel straight lines parallelogram perimeter perpendicular point in BC previous Ex quadrilateral rectangle required to prove respectively equal rhombus right angles right-angled triangle Standard Theorem straight line drawn trapezium triangle required Trisect vertex vertical angle
Populære avsnitt
Side 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.
Side 27 - If two triangles have two sides of the one equal to two sides of the...
Side 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.
Side 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.
Side 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.
Side 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...
Side 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.
Side 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.
Side 137 - ... upon the same side together equal to two right angles; the two straight lines shall be parallel to one another.
Side 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.