the fame Direction, that is, they are one and the fame ftrait Line. If this be denied, let CB, BE be both in the same strait Line; then will the Angle ABC + a ABE two right ones, = which is abfurd. b=ABC+ABD: a 13.3. XV. If two right Lines (AB, CD) mutually cut one another; the Angles (CEB, AED) at the Vertex fhall be equal. For the Angle AEC +CEB is d b hyp. cy ax. to two d 13. 3. right ones = AECAED; therefore CEB is AED. Q.E.D. SCHOLIUM E A D C G B H F I. If two right Lines (EA, AF) be drawn towards contrary parts from the Point A taken in any right Line GH, in fuch manner that the vertical Angles D and B are equal to one another; the faid Lines EA, AF, lie both in the fame strait Line. For the Angle D+A e 3 ax. two right Angles, f 13. 1. =B+A: therefore EA, AF, lie both in the g 14. 1. fame ftrait Line. Q. E. D. a from For because the Ang. AEC + AED+CEB DEB a four right Angles, therefore shall AECAED (= 1CEB + DEB) = two right b hyp. 2ax. Angles; and so `CED and AEB are right Lines. 15. I. 14. I. d 10. I. e 1 poft. f 3. I. g conft. i 4. I. B H I b Let the Lines AH, BE BC; in which continu'd and HI AH, and join out ACG. i h e = ther of the internal and oppofite Angles CAB, or CBA. bifect & the Sides AC, out take f EF BE, FC, IC, and continue ། Now because CE EA, and EF8EB, h15.1. and Angle FEC BEA; the Angle ECF fhall be EAB. In like manner, the Angle ICH therefore the whole Angle ACD BCG) is greater than either CAB Q. E. D. Side BC. Now because the Ang. m ACD+ACB CO 2. If a right Line (AE) makes C E D unequal Angles with another Line (CD) the Angle AED being acute, and AEC obtufe, the perpendicular Line AD let fall from any Point (A) thereof to that other Line CD, will fall next to the acute Angle AED. For if AC drawn towards the obtufe Angle be faid to be a Perpendicular; then in the Triangle AEC, the Angle AEC+ACE fhall be two right Angles, which is abfurd. 3. All the Angles of an equilateral Triangle, and those two of an Ifofceles one that are above the Bafe, are acute Angles. a. 17. I. From AC cut off ADb 3. 1. AB, and join DB; I. But & ADB d 16. 1. then the Angle ADB ABD. e gax. a 5. I. the Angle A be C; which is contrary to the Hypothefis: and if AB BC, then shall the b 18.1. Angle CA; which is contrary to the Hypo h 20.1. i 4 ax. thefis. Wherefore BC AB. And after the fame manner BC AC. Q.E. D. C XX. Two Sides (BA, AC) of every Triangle ABC, any ways taken are greater than the Side BC that remains. Continue out the Line BA, and take AD=AC, and draw the PROP. A D E B BC. XXI. If from the Extreme Points of one Side (BC) of a Triangle (ABC), two Lines ED, CD, be drawn to any Point within the Triangle; then are both thofe Lines taken together fhorter than the two other Sides of the Triangle. (BA, CA) but do contain a greater Angle BDC. C Produce BD to E, then is CE+ED=CD: add BD common to both, and then fhall BD+ DE ECCD+BD. Again, BA + AE BE; therefore BA+AC BE + EC. Whence, (1.) BA + ACBD + DC; which was one thing to be Demonftrated. (2.) The Ank 16.1. gle BDC DECA. Therefore the Angle BDCA. Q. E. D. PROP. PROP. XXII. To make a Triangle (FKG) of three right Lines FK, FG, GK, which shall be equal to three right Lines given A, B, C: but any two of them taken to gether must be longer than the third; because any two Sides of a Triangle taken together, are longer than the third. In the Infinite Line DE take orderly a DF, a 3. 1. FG, GH, equal to the given Lines A, B, C; then if with the Centers F and G, and the Diftances FD and GH, two Circles be barawn 3 post. cutting each other in K, and the right Lines KF, KG be join'd; the Triangle FKG fhall be d made, whofe Sides FK, FG, GK are equal to c 15 def. the three Lines DF, FG, GH, that is, to the d 1 ax. three given Lines A, B, C. Q: E. F. |