Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

Proposition XLVI.

IN Acute-angled Triangles(α) 1.1 the Square of any side (e. g. B. C, Fig. 87. N. 2.) subtending any of the Angles, as A is equal to the Squares of the other 2 sides (AB and AC) taken together, less 2 Rect∣angles (CAD) made by one side, containing the Acute-angle (CA) and its Segment AD reaching from the Acute-angle (A) to the Perpen∣dicular (BE) let fall from the other side.

Demonstration.

Make again BC=a, AC=b, AB=c, AD=x; then will CD=b−x. Therefore cc−xx= □ BD, and 〈 math 〉〈 math 〉 (i. e. □ BC−□CD) will also be = □ BD.

Therefore 〈 math 〉〈 math 〉.

i. e. (adding to both sides xx)

〈 math 〉〈 math 〉,

i. e. (adding on both sides bb, and subtracting 2bx)

〈 math 〉〈 math 〉. Q. E. D.

CONSECTARYS.

I. IF in the last Equation, except one, you add on both sides bb, and subtract aa, you'l have 〈 math 〉〈 math 〉, and, if moreover you divide hoth sides by 2b, you'l have 〈 math 〉〈 math 〉: Which is the Rule, having 3 sides given in an Acute-angled Triangle, to find the Segment AD, and consequently the Perpendicular BD.

Knowing therefore the Segments AD and CD, and also the Perpendicular BD in Oblique-angled Triangles, whether Ob∣tuse-angled or Acute-angled, when moreover the sides BC an•• AB are likewise given, the Angles of either Right-angled Tri∣angles or Oblique-angled ones, will be known; so that the la•••• Case of Plane Trigonometry, which we deferr'd from Prop. 34 to this place, may hence receive its solution.

Notes

• (α) 1.1

  Eucl. Prop. 13. l. 2.