## Elements of Geometry and Trigonometry |

### From inside the book

Results 1-5 of 100

Page 15

From the point C

From the point C

**draw**the line CF , making with AC the right angle ACF . Now , since ACD is a straight line , the angle FCD will be a right angle ( Prop . I. Cor . 1. ) ; and since ACE is a straight line , the angle FCE will likewise be ... Page 18

If from any point within a triangle , two straight lines be

If from any point within a triangle , two straight lines be

**drawn**to the extremities of either side , their sum will be less than the sum of the two other sides of the triangle . Let any point , as O , be taken within the trian- gle BAC ... Page 21

Then , take BD equal to AC , and

Then , take BD equal to AC , and

**draw**CD . Now , in the two triangles BDC , BAC , we have BD - AC , by construction ; the angle B equal to the angle ACB , by hypothesis ; B and the side BC common : therefore , the two triangles ... Page 22

From a given point , without a straight line , only one perpendicu lar can be

From a given point , without a straight line , only one perpendicu lar can be

**drawn**to that line . Let A be the point , and DE the given line .. B E Let us suppose that we can**draw**two perpendiculars , AB , AC . Page 23

Produce the perpendicular AB till BF is equal to AB , and

Produce the perpendicular AB till BF is equal to AB , and

**draw**FC , FD . First . The triangle BCF , is equal to the triangle BCA , for they have the right angle CBF CBA , the side CB common , and the - D side BF = BA ; hence the third ...### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Other editions - View all

### Common terms and phrases

ABCD adjacent altitude base become Book called centre chord circle circumference circumscribed common cone consequently construction contained corresponding cosine Cotang cylinder described diameter difference distance divided draw drawn equal equation equivalent evident expressed extremities fall figure follows formed formulas four frustum give given gles greater half hence homologous included inscribed intersection less likewise logarithm manner means measured meet middle multiplied opposite parallel parallelogram parallelopipedon pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment shown sides similar sine solid solid angle sphere spherical triangle square straight line Suppose surface taken tang tangent THEOREM third triangle triangle ABC vertex whole