## Plane Geometry |

### From inside the book

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Page 77

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**AC**, making a convenient with AB 2. Upon**AC**, lay off AD = DE = EF = FG : 3 . Draw HB . = GH . 4. Through D , E , F , G , and H , draw lines parallel to HB ,**meeting**AB at X , Y , Z , and W. Statement . AX = XY = YZ = ZW : WB . Proof ... Page 85

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**AC**. ( Prove AOC > ZACO . ) Ex . 187. Prove that the median to any side of a ...**AC**= DF ; In . A ABC , and △ DEF : BAC > ≤ D. BC > EF . Proof . 1 ...**meeting**BC at H. Draw GH . 4 . 5 . 6 . 7 . Note . A GAH △ ACH . ... GH = CH ... Page 91

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**AC**respectively of equilateral triangle ABC , such that AD = BE CF , then △ DEF is also equilat- eral . = F B C E ...**meeting**CA extended at E , then △ ADE is isosceles . LB. D Suggestion . - Compare E with C , and BDF with B F ... Page 122

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**meeting**at point 0 . 2 . Construct OM LAC . 3. With O as center and OM as radius , draw a O. Statement . This circle will be tangent to AB , BC , and**AC**. Proof . 1. O is equidistant from the sides of the triangle . § 169 2 . 3 . .. Is ... Page 128

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**meet**DE extended at B. 4. Construct**AC**, making / EAC = } ≤ A , and**meeting**DE extended at C. 5. Then △ ABC is the required triangle . Proof . 1. BAC = given ZA , since it equals 2 ( A ) . Const . 2. AD = 3. AE given ha and is an 128 ...### 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