## Elements of Geometry and Trigonometry |

### From inside the book

Results 1-5 of 39

Page 21

For , if these sides are not equal ,

For , if these sides are not equal ,

**suppose**AB to be the greater . 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 ... Page 22

B E Let us

B E Let us

**suppose**that we can draw two perpendiculars , AB , AC . Produce either of them , as AB , till BF is equal to AB , and D- draw FC . Then , the two triangles CAB , CBF , will be equal : for , the angles CBA , and CBF are right ... Page 23

Let us

Let us

**suppose**BC = BE ; then will the triangle CAB be equal to the the triangle BAE ; for BC = BE , the side AB is common , and the angle CBA - ABE ; hence the sides AC and AE are equal ( Prop . V. Cor . ) ... Page 24

Now , if it be possible ,

Now , if it be possible ,

**suppose**these two sides to be unequal , and that BC is the greater . On BC take BG = EF , and draw AG . Then , in the two triangles BAG , DEF , the angles R and E are equal , being right angles , the side BA ... Page 47

**Suppose**the chord AB- DE . Bisect these chords by the per- pendiculars CF , CG , and draw the radii CA , CD . D M A In the right angled triangles CAF , DCG , the hypothenuses CA , CD , are equal ; and the side AF , the half of AB ...### 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