The solution of geometrical exercises, explained and illustrated; with a complete key to the School Euclid1879 |
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Side 26
... of the oppo- site sides intersect in one point . 79. The exterior angles of a triangle are bisected , and the points of intersection of the bisectors is joined with the third angle ; shew that the third angle 26 THE SOLUTION OF.
... of the oppo- site sides intersect in one point . 79. The exterior angles of a triangle are bisected , and the points of intersection of the bisectors is joined with the third angle ; shew that the third angle 26 THE SOLUTION OF.
Side 30
... bisectors of two adjacent angles of a quadrilateral is half the sum of the other two angles ; and the angle contained by the bisectors of the two opposite angles is half the differ- ence of the other two angles . ( 32. ) 118. If the non ...
... bisectors of two adjacent angles of a quadrilateral is half the sum of the other two angles ; and the angle contained by the bisectors of the two opposite angles is half the differ- ence of the other two angles . ( 32. ) 118. If the non ...
Side 31
... bisected in E and F , shew that the angle EDF is equal to the angle BAC , and that AEDF is half the triangle . ( 38 ) . 124. A ... bisectors of the four interior or exterior angles of an unequal sided parallelogram enclose a rectangular ...
... bisected in E and F , shew that the angle EDF is equal to the angle BAC , and that AEDF is half the triangle . ( 38 ) . 124. A ... bisectors of the four interior or exterior angles of an unequal sided parallelogram enclose a rectangular ...
Side 32
... bisectors , CD , BE and AE . ( 38. ) 132. Through the angle A of the parallelogram ABCD a straight line AX is drawn upon which per- pendiculars are dropped from the other angles . Prove that the perpendicular from C is equal to the sum ...
... bisectors , CD , BE and AE . ( 38. ) 132. Through the angle A of the parallelogram ABCD a straight line AX is drawn upon which per- pendiculars are dropped from the other angles . Prove that the perpendicular from C is equal to the sum ...
Side 38
... bisectors of a triangle , to construct it . ( 34. ) 184. To find the locus of the middle points of all lines drawn from any point to a given straight line . ( 39. ) 185. To find the locus of a point , such that the sum of its distances ...
... bisectors of a triangle , to construct it . ( 34. ) 184. To find the locus of the middle points of all lines drawn from any point to a given straight line . ( 39. ) 185. To find the locus of a point , such that the sum of its distances ...
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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
ABC is greater ABCD Analysis angular points base BC bisect the base bisectors circle construct diagonal AC Draw the diagonal drawn parallel Drop the perpendicular equal to half equiangular equilateral triangle exterior angles four sides given line given point given straight line given triangle greater than AC half a right Hence hypotenuse intersect isosceles triangle joining the middle last exercise less Let AB Let ABC locus meeting BC middle point opposite angles opposite sides parallel to BC parallelogram pendicular perimeter PQRS produced proof Prop proposition quadrilateral radius rect rectangle contained rhombus right 2³ right 4º right angles right-angled triangle School Euclid Shew shewn equal Similarly solution square straight line drawn trapezium triangle ABC trisected vertex
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.