A Key to the Exercises and Examples Contained in a Text-book of Euclid's Elements: Books I.- VI. and XI.Macmillan, 1905 - 229 pages |
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Page 1
... triangle AFB is equilateral . 2. In Prop . 1 it has been shewn that AC and BC are each equal to AB ; and in Ex . 1 it has been shewn that AF and BF are each equal to AB . ... AC , BC , AF , BF are equal . Hence the fig . ACBF is a ...
... triangle AFB is equilateral . 2. In Prop . 1 it has been shewn that AC and BC are each equal to AB ; and in Ex . 1 it has been shewn that AF and BF are each equal to AB . ... AC , BC , AF , BF are equal . Hence the fig . ACBF is a ...
Page 2
... triangle : and AB = BC , by hypothesis ; ... BD BC , a radius of the circle CGH ; = .. the point D is on the circumference . 7. On AC describe an equilateral triangle ADC . With centre C , and radius CB , describe the circle BGH ...
... triangle : and AB = BC , by hypothesis ; ... BD BC , a radius of the circle CGH ; = .. the point D is on the circumference . 7. On AC describe an equilateral triangle ADC . With centre C , and radius CB , describe the circle BGH ...
Page 3
... triangle ABC , AC = AN , and BC = BM . [ Def . 15. ] Page 17 . 1. In the A3 DAB , CBA , DA = CB , being sides of a ... triangles ; .. the BCN = the A DCM in all respects ; ... BN DM . = [ I. 4. ] [ I. 4. ] [ L. 4. ] 3. In the △ BAY ...
... triangle ABC , AC = AN , and BC = BM . [ Def . 15. ] Page 17 . 1. In the A3 DAB , CBA , DA = CB , being sides of a ... triangles ; .. the BCN = the A DCM in all respects ; ... BN DM . = [ I. 4. ] [ I. 4. ] [ L. 4. ] 3. In the △ BAY ...
Page 4
... triangle ; and AY = AX , by hypothesis ; and the angle at A is common to both triangles ; ... BY CX . = 4. In the A ' BCD , ADC , Because- BC = AD , by hypothesis ; and DC is common to both ; and the BCD = the ADC , by hypothesis ...
... triangle ; and AY = AX , by hypothesis ; and the angle at A is common to both triangles ; ... BY CX . = 4. In the A ' BCD , ADC , Because- BC = AD , by hypothesis ; and DC is common to both ; and the BCD = the ADC , by hypothesis ...
Page 5
... triangles ; ... the DAB = the DCB . Similarly , by joining AC , it may be shewn that 2 . the ADC the ABC . = Join AC . Then in the △ ABC , CDA , ABCD , by hypothesis , Because and BC = DA , " " [ 1. 8. ] and AC is common to both ; 3 ...
... triangles ; ... the DAB = the DCB . Similarly , by joining AC , it may be shewn that 2 . the ADC the ABC . = Join AC . Then in the △ ABC , CDA , ABCD , by hypothesis , Because and BC = DA , " " [ 1. 8. ] and AC is common to both ; 3 ...
Other editions - View all
A Key to the Exercises and Examples Contained in a Text-Book of Euclid's ... H S 1848-1934 Hall No preview available - 2018 |
A Key to the Exercises and Examples Contained in a Text-Book of Euclid's ... H. S. Hall No preview available - 2017 |
Common terms and phrases
AB² ABCD AC² angles AP² arc AC AX² base BC BC² bisector bisects centre chord circum collinear common tangent concyclic constant cutting the given diagonals diam diameter diff draw drawn equiangular equilat fixed point given base given point given ratio given st greater half Hence identically equal inscribed intersect isosceles Join Let AB Let ABC Let AX locus middle point middle pt orthocentre par¹ parm pass pedal triangle perp plane polar produced Radical Axis radius equal rect respectively rhombus segment shewn sides Similarly solutions subtended sum of sqq tangent touch triangle vert vertex