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Make the side of the Triangle AB=a, BD=x, then will AD=2x. Since therefore the square BD i. e. xx being sub∣stracted out of the square AD i. e. 4xx, there remains the square AB 3xx, you'l have the Equation 〈 math 〉〈 math 〉; and dividing by 3 〈 math 〉〈 math 〉; therefore 〈 math 〉〈 math 〉
The Geometrical Construction. Having produced AB (n. 2.) to F a third part of it, the square of a mean proportional BD between BF and BA will be ⅓ aa or 〈 math 〉〈 math 〉, and so the Line BD = 〈 math 〉〈 math 〉. Therefore the Hypothenusa DA being divided in two in E, or at the interval BD, making the intersection from B and A, you'l have the Centre sought.
The Arithmetical Rule. Divide the square of the given side into three equal parts, and the square Root of a third part will give the semi-diameter AE or BE sought, by the intersection of two of which you'l have the Centre.