## Elements of Geometry: Containing the First Six Books of Euclid: With a Supplement ... to which are Added, Elements of Plane and Sphericale Trigonometry ... From the Last London Ed., Enl |

### From inside the book

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**ABC**is an equilateral**triangle**. Because the point A is the cen- tre of the circle BCD , AC is equal מן C A B E ( 11. Definition ) to AB ; and because the point B is the centre of the cir- cle ACE , BC is equal to AB : But it has been ... Page 13

... ABC to the triangle DEF ; and the other an- gles , to which the equal . sides are opposite , shall be equal , each to each , Да B E viz . the angle ABC to the angle DEF , and the angle ACB to DFE . For , if the

... ABC to the triangle DEF ; and the other an- gles , to which the equal . sides are opposite , shall be equal , each to each , Да B E viz . the angle ABC to the angle DEF , and the angle ACB to DFE . For , if the

**triangle ABC**be applied ... Page 14

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**triangle ABC**shall coincide with the whole triangle DEF , so that the spaces which they contain or their areas are equal ; and the remaining angles of the one shall coincide with the remaining angles of the other , and be equal to them ... Page 15

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**ABC**: And it has also been proved that the angle FBC is equal to the angle GCB , which are the angles upon the other side of the base . COROLLARY . Hence every equilateral**triangle**...**triangle**be equal to one another , the sides which ... Page 16

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**triangle**ACB ; produce AC , AD to E , F ; therefore , because AC is equal to AD in the**triangle**ACD , the angles ECD ...**ABC**, DEF be two**triangles**having the two sides AB , AC , equal to the two sides DE , DF , each to each , viz . AB ...### Other editions - View all

### Common terms and phrases

ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated described diameter divided draw equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given rectilineal given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROB PROP proportional proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle solid parallelopiped spherical angle spherical triangle square straight line BC THEOR touches the circle triangle ABC triangle DEF wherefore