## The solution of geometrical exercises, explained and illustrated; with a complete key to the School Euclid |

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The Solution of Geometrical Exercises, Explained and Illustrated: With a ... Charles Mansford Uten tilgangsbegrensning - 1875 |

The Solution of Geometrical Exercises, Explained and Illustrated Charles Mansford,Euclides Ingen forhåndsvisning tilgjengelig - 2015 |

### Vanlige uttrykk og setninger

ABCD Analysis base bisected bisectors centre circle common constant construct contained described diagonals difference distance divided draw Drop the perpendicular equal equilateral triangle exterior angles fall figure formed given line given point given straight line greater half Hence hypothesis interior intersect isosceles triangle Join known Lastly length less Let ABC line drawn locus meet method middle point opposite angles opposite sides parallel to BC parallelogram particular perimeter perpendicular PQRS problem produced proof Prop proposition proved quadrilateral radius rect rectangle remaining respects result rhombus right 28 right angles School Euclid segments Shew shewn Similarly solution square straight line third triangle ABC trisected twice units vertex vertical

### Populære avsnitt

Side 25 - 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.

Side 29 - Divide a given straight line into two parts such that the square on one of them may be double the square on the other.

Side 20 - In any triangle, if a straight line be drawn from the vertex to the middle of the base, twice the square of this line, together with twice the square of half the base, is equivalent to the sum of the squares of the other two sides of the triangle. Let...

Side 39 - Show that the locus of a point such that the sum of the squares of its distances from two fixed points is constant, is a circle.

Side 24 - ... exterior of two concentric circles, two straight lines be drawn touching the interior and meeting the exterior ; the distance between the points of contact will be half that between the points of intersection.

Side 31 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Side 22 - ... line is bisected at the middle point of the first. 37. If through any point equidistant from two parallel straight lines, two straight lines be drawn cutting the parallel straight lines, they will intercept equal portions of these parallel straight lines. 38. If the straight line bisecting the exterior angle of a triangle be parallel to the base, shew that the triangle is isosceles. 39. Find a point B in a given straight line CD, such that if AB be drawn to B from a given point A, the angle ABC...

Side 27 - Prove that the square on any straight line drawn from the vertex of an isosceles triangle to the base, is less than the square on a side of the triangle by the rectangle contained by the segments of the base : and conversely.

Side 29 - The straight line drawn from the vertex of a triangle to the middle point of the base is less than half the sum of the remaining sides.

Side 9 - The line which joins the middle points of two sides of a triangle is parallel to the third side, and is equal to half the third side.