Elementary geometry |
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Side 9
... bisector of an angle is the straight line that divides it into two equal angles . Def . 14. When one straight line stands upon another straight line and makes the adjacent angles equal , each of the angles is called a right angle . OBS ...
... bisector of an angle is the straight line that divides it into two equal angles . Def . 14. When one straight line stands upon another straight line and makes the adjacent angles equal , each of the angles is called a right angle . OBS ...
Side 11
... bisector . POSTULATES . Let it be granted that I. A straight line may be drawn from any one point to any other point . 2. A terminated straight line may be produced to any length in a straight line . 3. A circle may be described from ...
... bisector . POSTULATES . Let it be granted that I. A straight line may be drawn from any one point to any other point . 2. A terminated straight line may be produced to any length in a straight line . 3. A circle may be described from ...
Side 16
... bisectors of adjacent supplementary angles are at right angles to one another . 9. Find the angle between the bisectors ... bisector of the angle between the parts of the edge that meet at the crease will be at right angles to the crease ...
... bisectors of adjacent supplementary angles are at right angles to one another . 9. Find the angle between the bisectors ... bisector of the angle between the parts of the edge that meet at the crease will be at right angles to the crease ...
Side 20
... bisector of the angle BAC , meeting the base BC in X. Then in the triangles BAX , CAX we have BA = AC , AX common , ( Ax . 4. ) ( Hyp . ) and the included angle BAX = the included angle CAX . ( Hyp . ) Therefore the triangles are equal ...
... bisector of the angle BAC , meeting the base BC in X. Then in the triangles BAX , CAX we have BA = AC , AX common , ( Ax . 4. ) ( Hyp . ) and the included angle BAX = the included angle CAX . ( Hyp . ) Therefore the triangles are equal ...
Side 27
... DF , as DG ; and BC will fall as EG . ( Hyp . ) Let DH be the bisector of the angle FDG , ( Ax . 4. ) meeting EG in H. * Euclid , I. 21. + Euclid , I. 24 . Join FH . Then because in the triangles FDH , SECT . II . ] 27 TRIANGLES .
... DF , as DG ; and BC will fall as EG . ( Hyp . ) Let DH be the bisector of the angle FDG , ( Ax . 4. ) meeting EG in H. * Euclid , I. 21. + Euclid , I. 24 . Join FH . Then because in the triangles FDH , SECT . II . ] 27 TRIANGLES .
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Vanlige uttrykk og setninger
ABCD angle ABC angle APB angle BAC base bisect bisector Book called centre chord circle circumference coincide College common Constr construct converse Crown 8vo diagonals diameter difference distance divided double draw Edition ELEMENTARY equal angles Euclid Examples exterior angle extremities fall figure Find follows four Geometry given angle given point given straight line greater half Hence Illustrations included inscribed interior intersect isosceles triangle Join length less Let ABC line drawn locus magnitudes measure meet middle point minor arc opposite sides parallel parallelogram pass perpendicular placed polygon PROBLEM produced Professor Proof proportional quadrilateral radius ratio rectangle contained rectilineal figure regular required to prove respectively right angles School segment Shew sides similar Similarly square supplementary tangent THEOREM third touch triangle ABC unequal
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