## Elements of Geometry and Trigonometry |

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Page 19

Now , there may be three cases in the proposition , according as the point G falls without the triangle ABC , or upon its

Now , there may be three cases in the proposition , according as the point G falls without the triangle ABC , or upon its

**base**BC , or within it . First Case . The straight line GC < GI + IC , and the straight line AB < AI + IB ... Page 20

Let the side BA be equal to the side AC ; then will the angle C be equal to the angle B. A For , join the vertex A , and D the middle point of the

Let the side BA be equal to the side AC ; then will the angle C be equal to the angle B. A For , join the vertex A , and D the middle point of the

**base**BC . Then , the triangles BAD , DAC , will have all the sides of the one equal to ... Page 21

that side is generally assumed as the

that side is generally assumed as the

**base**, which is not equal to either of the other two . PROPOSITION XII . THEOREM . Conversely , if two angles of a triangle are equal , the sides oppo- site them are also equal , and the triangle is ... Page 30

... into as many triangles , less two , as the polygon has sides ; for , these triangles may be considered as having the point A for a common vertex , and for

... into as many triangles , less two , as the polygon has sides ; for , these triangles may be considered as having the point A for a common vertex , and for

**bases**, the several sides of the polygon , excepting the two sides which form ... Page 42

... divides the circle and its circumference into two equal parts . Let AEDF be a circle , and AB a diameter . Now , if the figure AEB be applied to AFB , their common

... divides the circle and its circumference into two equal parts . Let AEDF be a circle , and AB a diameter . Now , if the figure AEB be applied to AFB , their common

**base**AB retaining its position , the curve line AEB must fall ...### What people are saying - Write a review

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### 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