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Let the trapezium EBCF fall within the A DBC; let, also, the A DBC fall within the trapezium ABCD; and let all the figures stand on the same base BC: The perimeter of the A DBC is the perimeter of EBCF, and < the perimeter of ABCD.

First, let E and F be in the sides DB and DC of the A DBC, and let the vertex D of the ▲ DBC coincide with the A or the D of the trapezium ABCD.

Then, since (E. 20. 1.) DE+DF>EF, add to both, EB,BC, and CF; ... DE+EB+DF+FC +BC>EF+FC+CB+BE; i. e. the perimeter of the ▲ DBC > the perimeter of EBCF.

Again, since (E. 20, 1.) BA+AD > BD, add to both DC and CB; ... BA+AD+DC+CB> BD+DC+CB; i. e. the perimeter of the trapezium ABCD > the perimeter of the A DBC.

And, if E or F fall within the ▲ DBC, and the vertex of the A do not coincide with either of the A or D, of the trapezium, it may, in the same manner, be proved, that the proposition is true, a fortiori.


23. PROBLEM. One of the angles at the base of a triangle, the base itself, and the aggregate of the two remaining sides, being given, to construct the triangle.

Let K be the given angle, AB the given base

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of the triangle, and H the aggregate of the two remaining sides: It is required to construct the triangle.

At the point A, in AB, make (E. 23. 1.) the < BACK, and make (E. 3. 1.) AC=H; join C, B; and at the point B, in CB, make (E. 23. 1.) the CBD ACB: Then is DAB the triangle which was to be constructed. For, since (constr.) DCB= 2 DBC,


(E. 6. 1.) BD=DC; add to both DA; .. BD+DA=CD+DA; i. e. BD+DA=CA; and (constr.) CA=H; ... BD+DA=H; and the ZA was made equal to the given K: It is manifest, therefore, that DAB is the triangle which was to be constructed.


24. THEOREM. If two right-angled triangles have the three angles of the one equal to the three angles of the other, each to each, and if a side of the one be equal to the perpendicular let fall from the right angle upon the hypotenuse of the other, then shall a side of this latter triangle be equal to the hypotenuse of the former.

Let ACB and EDF be two right angled A,






right angled at C and D, having, also the ▲ DEF = 4 ABC, the EFD CAB, and the side AC, of the ▲ ABC, equal to the perpendicular DG, drawn D to the hypotenuse EF of the A DEF: The side DE, of the A

DEF, is equal to the hypotenuse AB, of the ▲



For, since AC=DG, and the two ABC, of the A ABC, are equal to the two , DGE, DEG, of the ▲ DEG, each to each, ... (E. 26. 1.) DE=AB.


25. THEOREM. If the sides of any given equilateral and equiangular figure of more than four sides, be produced so as to meet, the straight lines, joining their several intersections, shall contain an equilateral and equiangular figure, of the same number of sides as the given figure.

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angular figure, of more than four sides; let the sides, produced, meet in the points G, H, I, K, L, M; and let those points of intersection be

joined: Then is GHIKLM an equilateral and equiangular figure, of the same number of sides as ABCDEF.

For, since (hyp.) the A, B, C, D, E, F are all equal, the MAB, GBC, HCD, IDE, KEF, LFA are all (E. 13. 1. and E. 6. 1.) isosceles, and any two of them have their equal, each to each; ... since (hyp.) BA = AF, and that the MAB, MBA are equal to the LFA, LAF, each to each, the side MA of the ▲ MAB= the side LA E. 26. 1.) of the A LAF; and in the same manner it may be shewn that MB=GB, GC=CH, HD DI, IE=EK, and KF=FL: But, because the of the figure ABCDEF are (hyp.) equal, .*. (E. 15. 1.) the LAM, MBG, GCH, HDI, IEK, KFL, are all equal to one another; ... (E. 4. 1.) the sides LM, MG, GH, HI, IK and KL are all equal, as are also the of the ALAM, MBG, GCH, HDI, IEK, and KFL, each to each: And the AMB, BGC, CHD, DIE, EKF, and FLA have been shewn to be equal to one another: Wherefore the figure GHIKLM is equilateral and equiangular; and it is manifest that it has the same number of sides as the figure ABCDEF.


26. THEOREM. If two opposite sides of a quadrilateral figure be equal to one another, and the two

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