## Elements of Geometry and Trigonometry |

### From inside the book

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Page 26

**Opposite**exterior and interior angles lie on the same side of the secant line ... other within the parallels , but not adjacent : thus , EGB , GOD , are**opposite**exterior and interior angles ; and so also , are the angles AGE , GOC . Page 27

I. ) . Taking from each , the angle OGB , and there remains OGA = GOD . In the same manner we may prove that GOC = OGB . B D Cor . 3. If a straight line meet two parallel lines , the

I. ) . Taking from each , the angle OGB , and there remains OGA = GOD . In the same manner we may prove that GOC = OGB . B D Cor . 3. If a straight line meet two parallel lines , the

**opposite**exterior ... Page 29

Then , since AE , CB , are parallel , and CAD cuts them , the exterior angle DAE will be equal to its inte - C rior

Then , since AE , CB , are parallel , and CAD cuts them , the exterior angle DAE will be equal to its inte - C rior

**opposite**one ACB ( Prop . XX . Cor . 3. ) ; in like manner , since AE , CB , are parallel , and AB cuts them ... Page 30

In every triangle ABC , the exterior angle BAD is equal to the sum of the two interior

In every triangle ABC , the exterior angle BAD is equal to the sum of the two interior

**opposite**angles B and C. For , AE being parallel to BC , the part BAE is equal to the angle B , and the other part DAE is equal to the angle C. Page 32

In every parallelogram , the

In every parallelogram , the

**opposite**sides and angles are equal . Let ABCD be a parallelogram : then will AB = DC , AD = BC , A = C , and ADC = ABC . PROPOSITION XXVIII . THEOREM . B For , draw the diagonal BD .### 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 contained Cosine Cotang cylinder described determine diameter difference distance divided draw drawn equal equations equivalent expressed extremities faces feet figure follows formed four frustum give given gles greater half hence homologous hypothenuse included inscribed intersection less let fall logarithm manner means measured meet middle multiplied number of sides opposite parallel parallelogram pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quadrant quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment sides similar Sine solid solid angle sphere spherical triangle square straight line suppose taken Tang tangent THEOREM third triangle triangle ABC unit vertex whole