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And, if the angle Cee, which is common, be taken away, the remaining angle AEC will be equal to the remaining angle Bed. (Ax. 3.)
And, in the fame manner, it may be shewn that the angle aed is equal to the angle CIB. Q. E. D. COROLL.
All the angles made by any number of right lines, meeting in a point, are together equal to four right angles.
PRO P. XVI. THEOREM.
If one side of a triangle be produced, the outward angle will be greater than either of the inward opposite angles.
Let ABC be a triangle, having the fide AB produced to d; then will the outward angle CBD be greater than either of the inward opposite 'angles Bac or ACB.
For, bifect sc in E (Prop. 10.), and join Ae; in which, produced, take Ef equal to AE (Prop. 3.), and join BF.
Then, fince Ae is equal to EF, EC to EB (by conft.), and the angle AEC to the angle BEF (Prop. 15.), the angle Ace will, also, be equal to the angle EBF (Prop. 4.) But the angle cod is greater than the angle EBF; consequently it is also greater than the angle ace.
And, if co be produced to G, and AB be bisected, it may be shewn, in like manner, that the angle ABG, or its equal CBD, is greater than caB. Q. E. D.
PRO P. XVII. THEORE M.
The greater side of every triangle is opposite to the greater angle; and the greater angle to the greater side.
Let ABC be a triangle, having the side AB greater than the side Ac; then will the angle ACB be greater than the angle ABC.
For, since Ae is greater than Ac, let ad be taken equal to AC (Prop. 3.), and join cd.
Then, since cob is a triangle, the outward angle ADC is greater than the inward opposite angle DBC (Prop. 16.)
But the angle Acd is equal to the angle ADC, because Ac is equal to AD; consequently the angle Acd is, also, greater than Dec or ABC,
And, since Act is only a part of ACB, the whole angle. Acs must be much greater than the angle ABC.
Again, let the angle acB be greater than the angle ABC, then will the side AB be greater than the side Ac.
For, if Al be not greater than Ac, it must be either equal or less.
Bụt it cannot be equal, for then the angle ACB would be equal to the angle ABC (Prop. 5.), which it is not.
Neither can it be less, for then the angle Acb would be less than the angle ABC (Prop. 17.), which it is not.
The side AB, therefore, is neither equal to Ac, nor less than it; consequently it must be greater. Q.E.D.
PRO P. XVIII. THEOREM.
Any two sides of a triangle, taken together, are greater than the third fide.
Let ABC be a triangle; then will any two sides of it, taken together, be greater than the third fide.
For, in ac produced, take cd equal to CB (Prop. 3.), and join BD.
Then, because cd is equal to CB (by const.), the angle CDB will be equal to the angle CBD (Prop. 5.)
But the angle ABD is greater than the angle CBD, consequently it must also be greater than the angle ADB.
And, since the greater side of every triangle, is opposite to the greater angle (Prop. 17.), the fide ad is greater than the fide AB.
But ad is equal to Ac and ce taken together (by conft.); therefore AC, CB are also greater than AB.
And, in the same manner, it may be shewn, that any other two sides, taken together, are greater than the third fide. Q. E. D.
PRO P. XIX. PROBLEM.
To describe a triangle, whose sides shall be equal to three given right lines, provided any two of them, taken together, be greater than the third,
Let A, B, e be the three given right lines, any two of which, taken together, are greater than the third; it is required to make a triangle whose fides shall be equal to A, B, C respectively.
Draw any right line DG; on which take de equal to A, Ef equal to B, and FG equal to c (Prop. 3.)
From the point E, with the distance ED, describe the circle Khó, cutting og in K; and from the point F, with the distance FG, describe the circle GHL, cutting DG in L.
Then, because eg is greater than ed (by Hyp.), or its equal Ek, the point G, which is in the circumference of the circle GHL, will fall without the circle KHD.
And, because yd is greater than FG (by Hyp.), or its equa! Fļ, the point D, which is in the circumference of the circle KHD, will fall without the circle chl.
But fince a part of the circle GHL falls without the circle Khd, and a part of the circle Kid falls without the circle Ghl, neither of the circles can be included within the other.
Again, because DE, FG, or their equals EK, FL are, together, greater than EF (by Hyp.), the two circles can neither touch nor fall wholly without each other.
They must, therefore, cut one another, in some point H; and if the right lines EH, FH be drawn, EHF will be the triangle required.
For, since e is the centre of the circle KHD, Eh is equal to ED (Def. 13.) ; but ed is equal to A (by Const.); therefore en is also equal to A.
And, because F is the centre of the circle GHL, FH is equal to FG (Def. 13.) ; but rg is equal to c (by Conf.); therefore rh is also equal to c.
And fince Er is, likewise, equal to B (by Const.), the three sides of the triangle EHF are respectively equal to the three given lines A, B, C, which was to be shewn.