## Plane Geometry |

### From inside the book

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**Bisector**of an angle . ( j ) Right angle ; acute ; obtuse ; straight . ( k ) Complementary angles ; supple- mentary ; adjacent ; vertical . ( 1 )**Perpendicular**lines . Ex . 119. What is an axiom ? a theorem ? h Ex . 120 . What is the ... Page 42

... perpendicular to a segment at its mid - point is the

... perpendicular to a segment at its mid - point is the

**Perpendicular**-**bisector**of the segment . Ex . 63. Draw a line BC , 3 in . in length . With compasses , locate a point A above BC which is 2 in . from B and 2 in . from C. Locate simi ... Page 43

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**perpendicular**-**bisector**of the segment . 12 E B Hypothesis . C and D are equidistant from the ends of seg- ment AB . CD intersects AB at E. Conclusion . Proof . 2 . 3 . 4 . 2i + 5 . 6 . 7 . AE = EB ; CDL AB . 1. In △ ACD and A CDB ... Page 44

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**perpendicular**-**bisector**of AB . [ If two points are each equidistant from the ends of a segment , they determine the**perpendicular**-**bisector**of the segment . ] $ 17 $ 77 Ex . 65. Divide a given segment into four equal parts . Ex . 66 ... Page 45

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**perpendicular**-**bisector**of DE . Why ? 5 . .. CFL AB at C. [ Since AB and DE are the same straight line . ] 81. Proposition X proves that one perpendicular can be drawn to a line at a point in the line . It can be proved that only one ...### 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