Text-book of Elementary Plane Geometry |
Other editions - View all
Common terms and phrases
abc2 ABCD adjacent angles adjacent side altitude angle contained angle equal angular points base bisecting the angle centre of similitude chord circle touches circles cut circumference coincide containing the right corresponding lines corresponding points diagonals diameter draw drawn parallel equilateral triangle given angle given circles given figure given line given point given side half the perimeter hypothenuse inscribed circle isosceles triangle legs line bisecting line is drawn line joining line of centres mean proportional middle point number of sides opposite angle opposite side parallelogram pendicular perimeter perpendicular point of intersection points of contact points of division potense Prove quadrilateral radii ratio rectangle regular polygon required line rhombus right-angled triangle secant semicircle Shew sided figure sides containing similar square straight line tangent thereupon third side trapezium triangle ABC triangles are congruent vertex vertical angle
Popular passages
Page 13 - In a right triangle, the perpendicular from the vertex of the right angle to the hypotenuse is a mean proportional between the segments of the hypotenuse: p2 = mn. Any two similar figures, in the plane or in space, can be placed in " perspective," that is, so that lines joining corresponding points of the two figures will pass through a common point.
Page 52 - The diagonals of a quadrilateral intersect at right angles. Prove that the sum of the squares on one pair of opposite sides is equal to the sum of the squares on the other pair.
Page 25 - Each side of a triangle is smaller than the sum of the other two, and greater than their difference. The first part of this theorem is an immediate consequence of the Axiom of Distance (54) ; that is, AC < AB + BC. Subtract AB from both members of this inequality, and AC — AB < BC. That is, BC is greater than the difference of the other sides. Prove the same for each of the other sides.
Page 10 - If a straight line, fatting on two other straight lines, make an exterior angle equal to the interior opposite angle on the same side of the...
Page 50 - The area of any polygon circumscribing a circle is equal to half the product of the radius of the circle, and the perimeter of the polygon. (Divide the polygon into triangles, with the centre for vertex.) tEx.
Page 46 - If two circles touch each other, and also touch a given straight line, the part of the straight line between the points of contact is a mean proportional between the diameters of the circles.
Page 14 - Ьs + cs - 2Ьc cos A, and apply it to prove that if the straight line which bisects the vertical angle of a triangle also bisects the base, then the triangle must be isosceles. 9. Find the area of a triangle in terms of the sides. 10. Find the radius of the circle which touches one side of a triangle and the two other sides produced.
Page 41 - ... as any homologous altitudes. 549. EXERCISE. The perimeters of similar triangles are to each other as any homologous medians. 550. EXERCISE. The perimeters of two similar polygons are 78 and 65 ; a side of the first is 9, find the homologous side of the second. 551. DEFINITION. A line is divided in extreme and mean ratio when it is divided into two parts so that one segment is a mean proportional between the whole line and the other segment. PROPOSITION XXVIII. PROBLEM 552. To divide a line in...
Page 32 - From any point in the base of an isosceles triangle perpendiculars are drawn to the sides ; prove their sum to be equal to the perpendicular drawn from either basal vertex to the opposite side.
Page 45 - The lengths of the circumferences of two circles are to each other as the radii.