Let BD be produced to E. Then is CE + ED a a 20. 1. CD, and BD common to both, b then ſhall be BD DE+ EC CD + BD. Again, BA AE b 4. ax.a BE. b therefore BA AC BE + EC. Wherefore 1. BA + AC BD + DC. 2. The angle BDC cc 16. 1. DEC c A. Therefore the angle BDC A. Which was to be demonſtrated.

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To make a triangle FKG of three right lines FK, FG, GK, which ſhall be equal to thee right lines given A, B, C. Of which it is neceſſary that any two taken together be longer than the third.

From the infinite line DE a take DF, FG, GH a 3. 1.equal to the lines given, A, B, C. Then it from the b centers F and G by the diſtances of FD b 3.poſt.and GH, two circles be drawn cutting each other in K, and the right lines KF, KG be joined,c 15. def. the triangle FKG ſhall be made, c whoſe ſides FK, FG, GK are equal to the three lines d 1. ax.DF, FG, GH d that is to the three lines given A, B, C. Which was to be done.

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a 1. poſt. At a point A in a right line given AB to make a right-lined angle A equal to a right-lined angle given D.