## Plane Geometry |

### From inside the book

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**radii**can be made to coincide and are called equal circles . Hence : All**radii**of the same circle or of equal circles are equal . Ex . 31 . Draw a circle of radius 1 inch . Ex . 32. Draw two circles having the same center with**radii**... Page 23

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**radii**of the same circle or of equal circles are equal . ( § 17. ) 3. A straight angle equals two right angles . ( § 32. ) 4. All straight angles are equal . ( § 33. ) 5. The sum of all the successive adjacent angles around a point on ... Page 41

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**radii**, draw arcs intersecting at E. 3 . Statement . Draw OE . OE bisects AOB . Proof . The proof is to be given by the pupil . Suggestions . Draw CE and DE . Recall § 66 . Note . - For construction problems , the regular form is to ... Page 44

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**radii**, draw arcs intersecting at C and also at D. 2 . Statement . Draw CD intersecting AB at E. E bisects AB . Proof . 1. AC = BC , and also AD = BD . [**Radii**of equal circles are equal . ] 2 . .. CD is the perpendicular - bisector ... Page 46

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**radii**, draw two arcs intersecting at F. 3 . Statement . Draw CF. CFLAB . Proof . 1 . 2 . C is equidistant from D and E. Fis equidistant from D and E. Why ? Why ? 3 . .. CF is the perpendicular - bisector of DE . Why ? 4 . .. CFL AB ...### Other editions - View all

### Common terms and phrases

ABCD acute angle adjacent angles adjoining figure altitude angles are equal apothem base bisector bisects central angle chord circle of radius circumscribed polygons Conclusion congruent Construct a triangle Determine diagonals diameter divide Draw drawn equal angles equal circles equal respectively equal sides equidistant equilateral triangle extended exterior angle geometry given circle given point given segment given triangle Hence homologous sides hypotenuse Hypothesis intersect isosceles trapezoid isosceles triangle length mean proportional median meeting AC mid-point Note number of sides opposite sides parallel parallelogram pentagon perigon perimeter perpendicular perpendicular-bisector PROPOSITION quadrilateral radii ratio rectangle regular inscribed polygons regular polygon rhombus right angle right triangle secant similar triangles straight angle straight line Suggestion Suggestions.-1 Supplementary Exercises tangent THEOREM trapezoid triangle ABC triangle equal Try to prove vertex ZAOB

### Popular passages

Page 166 - The sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse.

Page 207 - The areas of two similar triangles are to each other as the squares of any two homologous sides.

Page 166 - The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.

Page 83 - If two sides of a triangle are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side.

Page 170 - If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar.

Page 105 - A tangent to a circle is perpendicular to the radius drawn to the point of contact.

Page 86 - If two triangles have two sides of one equal, respectively, to two sides of the other...

Page 204 - The formula states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the base and altitude.

Page 299 - Prove that an equiangular polygon inscribed in a circle is regular if the number of sides is odd. Ex.

Page 194 - Two rectangles are to each other as the products of their bases and altitudes. For if R = a6, and R