 | Archibald Patoun - 1734 - 568 sider
...the Parallelograms BKLH and KCML-, but the Sum of thefe Parallelograms is equal to the Square B CMH, therefore the Sum of the Squares on AB and AC is equal to the Square on BC. Cor. i. Hence in a rightangled Triangle, the Hypothenufe and one of the Legs being given, we may eafily... | |
 | Robert Gibson - 1806 - 486 sider
...ACGF, is equal to the parallelogram KCLM. So ABDE + ACGF the sum of the Plate I. squares = BELH + KCML, the sum of the two parallelograms or square BCMH ;...squares on AB and AC is equal to the square on BC. QED Cor. Hence the hypothenuse of a right-angled triangle may be found by having the legs ; thus, the... | |
 | Robert Gibson - 1808 - 482 sider
...be proved, that the square ACGF, is equal to the parallelogram KCLM. So ABDE + ACGF the sum of the Plate I. squares, = BKLH+KCML, the sum of the two...squares on AB and AC is equal to the square on BC. -QED Cor. 1. Hence the hypothenuse of a rightangled triangle may be found by having the legs ; thus,... | |
 | Robert Gibson - 1811 - 580 sider
...square ACGF, is equal to the parallelogram KCLM. So AB DE + ACGF the sum of the &quvces=BKLH+KCML, the sum of the two parallelograms or square BCMH ;...squares on AB and AC is equal to the square on BC. 2. ED Cor. 1. Hence the hypothenuse of a right-angled triangle may Its found by having the sides ;... | |
 | Robert Gibson - 1814 - 558 sider
...ABDE+ACGFûie sum of the squares= BKLH + KCML, the sum of the two parallelograms or square BCMH ; thei'efore the sum of the squares on AB and AC is equal to tiie square on BC. QED Cor. 1. Hence the hypothenuse of a right-angled triangle may be found by having... | |
 | Robert Gibson - 1821 - 594 sider
...may be proved, that the square ACGF, is equal to the parallelogram KCLM So ABDE+ACGF the sum of the squares =* BKLH+KCML, the sum of the two parallelograms...square BCMH ; therefore the sum of the squares on A Band AC is equal to the square on BC. QE D* Cor. 1. Hence the hypothenuse of a right-angled triangle... | |
 | Nathaniel Bowditch - 1826 - 764 sider
...parallelograms BKLH and KCML : the sum of these parallelograms is equal to the square BCMH, therefore sum of the squares on AB and AC is equal to the square on BC. Cor. Hence in any right angled triangle, if we have the hypotenuse one of the legs, we may easily find... | |
 | Nathaniel Bowditch - 1826 - 732 sider
...BKLH and KCML ; but lhe. sum of these parallelograms is equal to the square BCMH, therefore the iyirn of the squares on AB and AC is equal to the square on BC. Cor. Hence in any right angled triangle, if we have the hypotenuse and one of the legs, we may easily... | |
 | Robert Gibson - 1832 - 290 sider
...may be proved that the square ACGF is equal to the parallelogram KCLM. So ABDE+ACGF the sum of the squares —BKLH-\-KCML, the sum of the two parallelograms...squares on AB and AC is equal to the square on BC. QED* . Cor. 1. Hence the hypothenuse of a right-angled triangle may be found by having the sides :... | |
 | Robert Gibson, James Ryan - 1839 - 452 sider
...may be proved that the square ACGF is equal to the parallelogram KCLM. So ABDE+ACGF the sum of the squares =BKLH+KCML, the sum of the two parallelograms or square BCMH; therefore the sum of ihe squares on AB and AC is equal to the square on BC. QED* Cor. 1. Hence the hypothenuso of a right-angled... | |
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KCML, the sum of the two parallelograms or square BCMH ; therefore the sum of the squares on AB and AC is equal to the square on BC. 
