## Elements of Plane Geometry: For the Use of Schools |

### From inside the book

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**draw**from the point C , where the two**lines**A Fig . 1 . E D C B meet , the**line**CE perpendicular to AB ; the angles ACE , ECB , will be right angles ( Def . 6 ) , and their sum , consequently , two right angles ; but the opening between the ... Page 15

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**Draw**from B the**line**BD , so as to divide the angle B into two equal parts ; then in the two triangles ABD , CBD , we have the two sides AB , BD , and the included angle ABD , in the one , respectively equal to the two sides BC , BD ... Page 16

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**draw the line**BD ; then we have the side AD of the triangle ABD equal to the side BC of the triangle ACB , the side AB common to both triangles , and the angle DAB of one equal to the angle CBA of the other ( by hypothesis ) , hence the ... Page 18

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**lines**are parallel . Let the angles AIF , EOD , be equal ; then we have to prove that AB , CD , are parallel . If they are not Fig . 11 . parallel ,**draw**through I , the**line**HK , par- allel to CD ; then will A HIF be equal to EOD ... Page 19

... Draw IH perpendicular to AB , it will also ( by Prop . 9 , Cor . 2 ) be perpendicular to both CD and EF ; hence these lines , being both perpendicular to the same line ...

... Draw IH perpendicular to AB , it will also ( by Prop . 9 , Cor . 2 ) be perpendicular to both CD and EF ; hence these lines , being both perpendicular to the same line ...

**draw the line**BE parallel to AC ; then will the angles EBD , BAC ...### Other editions - View all

Elements of Plane Geometry: For the Use of Schools - Primary Source Edition Nicholas Tillinghast No preview available - 2013 |

### Common terms and phrases

ABCD adjacent angles allel alternate angles altitude angles ABD angles is equal antecedent and consequent B. I. Ax centre circle whose radius circumference circumscribed circumscribed circle common measure Converse of Prop describe an arc diameter divided draw the line equal angles equal B. I. Prop equal chords equal Prop equal respectively equally distant equiangular equivalent feet four numbers given angle given line given point given side half hence the triangles hypotenuse included angle inscribed angle Let ABC linear units longer than AC multiplied number of sides number of square oblique lines opposite parallel parallelogram perimeter perpendicular PROBLEM prove quadrilateral radii rectangle regular polygons respectively equal right angles Prop right-angled triangle Scholium sides AC similar subtended tangent THEOREM three sides triangle ABC triangles are equal vertex

### Popular passages

Page 31 - A circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within called the center.

Page 63 - The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides.

Page 71 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.

Page 53 - In any proportion, the product of the means is equal to the product of the extremes.

Page 89 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.

Page 54 - In a series of equal ratios, any antecedent is to its consequent, as the sum of all the antecedents is to the sum of all the consequents. Let a: 6 = c: d = e :/. Then, by Art.

Page 83 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Page 59 - The area of a parallelogram is equal to the product of its base and its height: A = bx h.

Page 16 - Conversely, if two angles of a triangle are equal, the sides opposite them are also equal, and the triangle is isosceles.

Page 61 - From this proposition it is evident, that the square described on the difference of two lines is equivalent to the sum of the squares described on the lines respectively, minus twice the rectangle contained by the lines.