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IN an obtuse angled Triangle having the thr••e sides given to find the Perpendicular let fall from the Vertex to the Base being continued: i. e. Having given AB, BC, AC (Fig. 22. n. 1.) to find AD or CD (for the one being found the other will readily be so also) Coroll Prop. 12. Lib. 2. Eucl.
What we premonished about the former Problem, we un∣derstand to be premonish'd here also. For the rest make here also AB=a, BC=b, AC=c, CD=x; then will BD= b+x: Wherefore by the Pythagorick Theorem the □ AD will = cc−xx, and by the same Theorem the same □ AD = 〈 math 〉〈 math 〉.
Therefore 〈 math 〉〈 math 〉; and adding to both sides 〈 math 〉〈 math 〉; and transposing cc and 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉.
The Arithmetical Rule. Substract from the square of the greater side the sum of the squares of the base and lesser side; and the Remainder divided by double the base will give its continuation to the Perpendicular.
Geometrical Construction from the Equation reduc'd: Having described a semi-circle upon AB (n. 2.) apply therein AC, and the □ of CB drawn will 〈 math 〉〈 math 〉; which since it must be divided by 2b make, as CF=2b to CE 〈 math 〉〈 math 〉, so CE to CD the segment sought, n 1.