| Archibald Patoun - 1734 - 414 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 - 1795 - 319 sider
...Sum of the Squares, Plate I. Squares, = BKLH+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.** Q^ ED Cor. i. Hence the Hypothenufe of a rightangled Triangle may be found by having the Legs ; thus,... | |
| Robert Gibson - 1806 - 452 sider
...ACGF the sum of the Plate I. squares = BELH + 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.** QED Cor. Hence the hypothenuse of a right-angled triangle may be found by having the legs ; thus, the... | |
| Robert Gibson - 1808 - 440 sider
...ACGF the sum of the Plate I. squares, = BKLH+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.** -QED Cor. 1. Hence the hypothenuse of a rightangled triangle may be found by having the legs ; thus,... | |
| Robert Gibson - 1811 - 508 sider
...So AB DE + ACGF the sum of the &quvces=BKLH+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.** 2. ED Cor. 1. Hence the hypothenuse of a right-angled triangle may Its found by having the sides ;... | |
| Robert Gibson - 1814 - 508 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... | |
| Nathaniel Bowditch - 1826 - 617 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 the other leg,... | |
| Nathaniel Bowditch - 1826 - 617 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 find the other... | |
| Robert Gibson - 1832 - 348 sider
...ABDE+ACGF the sum of the squares —BKLH-\-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.** QED* . Cor. 1. Hence the hypothenuse of a right-angled triangle may be found by having the sides :... | |
| Robert Gibson, James Ryan - 1839 - 412 sider
...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 triangle may be found by having the sides i thus,... | |
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