Elementary Geometry ... |
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Página 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 ...
Página 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 ...
Página 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 ...
Página 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 ...
Página 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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Términos y frases comunes
AB² ABCD alternate angles angle ABC angle APB angle BAC angles are equal angles equal bisector centre chord circumference coincide College Constr construct contrapositive Crown 8vo describe a circle diagonals diameter divided draw Edition equal angles equal circles equiangular Euclid exterior angle fcap Find the locus four right angles Geometry given angle given circle given line given point given rectilineal figure given straight line hypotenuse included angle inscribed intersect isosceles triangle less Let AB Let ABC locus magnitudes meet middle point minor arc opposite angles opposite sides parallel parallelogram perpendicular polygon PROBLEM produced Proof Q. E. D. THEOREM quadrilateral radius rectangle contained reflex angle required to prove rhombus right angles segment semicircle Shew square subtend supplementary supplementary angles tangent THEOREM trapezium triangle ABC triangles are equal unequal
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