CHAP. XIX. To finde the perpendicular of an equilater Tri∣angle, the sides thereof being given.
AN equilater Triangle is that, which hath all his three sides equall; for I neede not stand to define what a Triangle is, seeing it is so well knowne to be a figure made of three lines, either Right or Sphericall, but Page 33 here I speak of Right lined Triangles. Now to finde the length of the perpendicular, which is a line falling from the opposite angle to the base, making Right Angles therewith; lay the threed upon the side of the Quadrat, and rectifie the bead to the length of one of the sides; then opening the threed, untill the bead fall upon the right parallel of halfe the base; so shall it shew the con∣trary parallel of the perpendicular.
Let the Triangle ABC, be given, wherein it is re∣quired to since the length of the perpendicular AD: First, lay the threed upon the side of the Quadrat, and rectifie the bead to 8 the length of the side AB, or AC; then open the threed, untill the bead falleth upon the Right parallel of 4, the one halfe of the base BC, which is BD, or CD; so shall the bead cut the contrary pa∣rallel of 6 9/10 which is the length of the perpendicular AD, which was required.
Thus may you likewise finde the perpendicular of an Isoscheles Triangle, which is a Triangle of two equall sides; for, first placing the bead to one of the sides, then Page 34 opening the threed untill the bead fall on the Right paral∣lel of one halfe of the base; so shall it shew the contrary parallel of the perpendicular.