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
... Hence so that the 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 ...
... Hence so that the 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 ...
Page 4
... Hence , adding the equal angles in ( i ) and ( ii ) , the ABC the ADC . = 2. In the ABC , since AB = AC , by ... Hence the whole ABD = the whole ACD . 3. As in the last example it may be shewn that the and the .. the remaining ABC the ...
... Hence , adding the equal angles in ( i ) and ( ii ) , the ABC the ADC . = 2. In the ABC , since AB = AC , by ... Hence the whole ABD = the whole ACD . 3. As in the last example it may be shewn that the and the .. the remaining ABC the ...
Page 6
... Hence the sq . ABCD coincides with the sq . EFGH , and is therefore equal to it in area . 13. Join BD , FH ; and apply the fig . ABCD to the fig . EFGH , so that A falls on E , and AB falls along EF ; then B must fall on F , for AB EF ...
... Hence the sq . ABCD coincides with the sq . EFGH , and is therefore equal to it in area . 13. Join BD , FH ; and apply the fig . ABCD to the fig . EFGH , so that A falls on E , and AB falls along EF ; then B must fall on F , for AB EF ...
Page 12
... Hence the AOX , COY are complementary . 4. The COX , XOA are supplementary ; and the BOX = the XOA ; ... the BOX , COX are supplementary . The second case is similarly proved . Page 36 . 1. In the CDA , because DA = DC , .. the DAC ...
... Hence the AOX , COY are complementary . 4. The COX , XOA are supplementary ; and the BOX = the XOA ; ... the BOX , COX are supplementary . The second case is similarly proved . Page 36 . 1. In the CDA , because DA = DC , .. the DAC ...
Page 23
... Hence AB , CD are equal and may be shewn parallel . fig . ABCD is a par TM . 7 . [ 1. 15. ] [ 1. 26. ] and I. 4 ] . Hence the [ I. 33. ] - △ ABC = < ACB , being halves of equal < 3 . ... AB AC . 8. In △ ABD , BDC , Because AB = DC ...
... Hence AB , CD are equal and may be shewn parallel . fig . ABCD is a par TM . 7 . [ 1. 15. ] [ 1. 26. ] and I. 4 ] . Hence the [ I. 33. ] - △ ABC = < ACB , being halves of equal < 3 . ... AB AC . 8. In △ ABD , BDC , Because AB = DC ...
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