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Side 61
... intersecting the former circle on the side of AC remote from B in D. Join AD , CD , and BD ; BD bisects the angle ABC . Proof . In the triangles ABD , CBD , because and BD is common , AB = BC , ( Constr . ) and the base AD = the base DC ...
... intersecting the former circle on the side of AC remote from B in D. Join AD , CD , and BD ; BD bisects the angle ABC . Proof . In the triangles ABD , CBD , because and BD is common , AB = BC , ( Constr . ) and the base AD = the base DC ...
Side 63
... with centre B and the same radius describe a circle intersecting the former in two points C and D. Join CD cutting AB in O. * Euclid , 1. 12 . + Euclid , 1. 10 . Then will be the point of bisection . Join AD SECT . IV . ] 63 PROBLEMS .
... with centre B and the same radius describe a circle intersecting the former in two points C and D. Join CD cutting AB in O. * Euclid , 1. 12 . + Euclid , 1. 10 . Then will be the point of bisection . Join AD SECT . IV . ] 63 PROBLEMS .
Side 64
... intersect in R. Join RP , RQ . RPQ is the triangle required . Proof . For RPQ has its three sides respectively equal to A , B and C * . Euclid , I. 22 . - Limitation . It is necessary that any two of 64 [ BOOK I. ELEMENTARY GEOMETRY .
... intersect in R. Join RP , RQ . RPQ is the triangle required . Proof . For RPQ has its three sides respectively equal to A , B and C * . Euclid , I. 22 . - Limitation . It is necessary that any two of 64 [ BOOK I. ELEMENTARY GEOMETRY .
Side 73
... intersecting straight lines is the pair of lines , at right angles to one another , which bisect the angles made by the given lines . Let AB , DC intersect in O ; it is required to find a point equally distant from AB and DC . A N P ...
... intersecting straight lines is the pair of lines , at right angles to one another , which bisect the angles made by the given lines . Let AB , DC intersect in O ; it is required to find a point equally distant from AB and DC . A N P ...
Side 74
... intersect , is the bisectors of the angles between the lines . It may further be proved that no point not in a bisector is equally distant from these lines ; that is , the bisectors are the complete locus . EXERCISES . Find the ...
... intersect , is the bisectors of the angles between the lines . It may further be proved that no point not in a bisector is equally distant from these lines ; that is , the bisectors are the complete locus . EXERCISES . Find the ...
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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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