Plane Geometry |
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Page 151
... chords , AC and BC , of the arc . Draw the perpendicular bisectors of these chords . How is the center of the circle obtained ? The radius ? A C B 3. From the measurements of a piece of a broken wheel a new wheel is to be cast of the ...
... chords , AC and BC , of the arc . Draw the perpendicular bisectors of these chords . How is the center of the circle obtained ? The radius ? A C B 3. From the measurements of a piece of a broken wheel a new wheel is to be cast of the ...
Page 152
... chord bisects the chord and the subtended arcs . A C E B Ф O Hypothesis . AB is a chord of a circle with center 0 . Line CD passes through O , intersecting the circle at Cand D. And CD ⊥ AB at E. Conclusion . AE = BE , AC = BC , and AD ...
... chord bisects the chord and the subtended arcs . A C E B Ф O Hypothesis . AB is a chord of a circle with center 0 . Line CD passes through O , intersecting the circle at Cand D. And CD ⊥ AB at E. Conclusion . AE = BE , AC = BC , and AD ...
Page 153
... chord is perpendicular to the chord . 2. A radius which bisects an arc bisects the chord which subtends it and is perpendicular to the chord . 3. A line which bisects a chord and also its subtended arc passes through the center of the ...
... chord is perpendicular to the chord . 2. A radius which bisects an arc bisects the chord which subtends it and is perpendicular to the chord . 3. A line which bisects a chord and also its subtended arc passes through the center of the ...
Page 154
... chords are equidistant from the center . C N D O B A M Hypothesis . In circle with center O , chords AB and CD are equal , and OMLAB and ONI CD . Conclusion . OM = ON . Suggestions . For proving OM = ON , what auxiliary lines must be ...
... chords are equidistant from the center . C N D O B A M Hypothesis . In circle with center O , chords AB and CD are equal , and OMLAB and ONI CD . Conclusion . OM = ON . Suggestions . For proving OM = ON , what auxiliary lines must be ...
Page 155
... chords are unequally distant from the center , the greater chord being at the less distance . AK E C P 2 14 13 0 M B M D Hypothesis . In circle with center O , AB and CD are chords , AB > CD , OM ⊥ AB , and ONI CD . Conclusion . OM ...
... chords are unequally distant from the center , the greater chord being at the less distance . AK E C P 2 14 13 0 M B M D Hypothesis . In circle with center O , AB and CD are chords , AB > CD , OM ⊥ AB , and ONI CD . Conclusion . OM ...
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Common terms and phrases
ABCD AC and BC acute angle altitude angle equal angles are equal apothem base bisects central angle chord circle with center circumscribed compasses and straightedge Conclusion congruent Construct a triangle Corollary corresponding sides diagonals diameter distance Divide a given drawn equal angles equal arcs equal circles equilateral triangle EXERCISES exterior angles follows formed geometry given angle given circle given line given line-segment given point given triangle Hence hypotenuse Hypothesis inscribed angle intercepted internally tangent isosceles triangle locus measure medians middle points number of sides parallel lines parallelogram perimeter perpendicular bisector point of contact proof in full proof is left quadrilateral radii radius ratio rectangle regular polygon rhombus right angle right triangle secant segment Show similar polygons straight angle straight line student SUGGESTION tangent trapezoid triangle are equal vertex Write the proof Нур
Popular passages
Page 225 - ... they have an angle of one equal to an angle of the other and the including sides are proportional; (c) their sides are respectively proportional.
Page 76 - 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 46 INTERCEPTS BY PARALLEL LINES.
Page 132 - If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second.
Page 53 - Two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. 2. Two triangles are congruent if two angles and the included side of one are equal respectively to two angles and the included side of the other.
Page 4 - PERIPHERY of a circle is its entire bounding line ; or it is a curved line, all points of which are equally distant from a point within called the center.
Page 72 - There are three important theorems in geometry stating the conditions under which two triangles are congruent: 1. Two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other.
Page 260 - S' denote the areas of two circles, R and R' their radii, and D and D' their diameters. Then, I . 5*1 = =»!. That is, the areas of two circles are to each other as the squares of their radii, or as the squares of their diameters.
Page 199 - In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides, minus twice the product of one of these sides and the projection of the other side upon it.
Page 203 - In any quadrilateral the sum of the squares of the four sides is equal to the sum of the squares of the diagonals, plus four times the square of the line joining the middle points of the diagonals.
Page 47 - In a right triangle, the side opposite the right angle is called the hypotenuse and is the longest side.