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Again, :: G is the centre of O HLK,.. GH=GK;
(def. 15) but GH=CR, (constr.) :. GK = CR; and FG = BQ;
(constr.) :. the three st. lines KF, FG, GK, are respectively=the three AP, BQ, CR:
.. A KFG has its three sides equal to the three given lines.
Q. E. F.
['but any two whatever of these must be greater than the third.' Otherwise no triangle could be formed, as has been shown in prop. 20. The beginner may, with ruler and compasses, draw diagrams for the cases where the sum of two of the lines is equal to, and when it is less than the third line, and they will exhibit the impossibility of making a triangle.]
From a given point draw three lines of given lengths, so that their extremities may be in one line and equally distant from each other.
When is this not possible ?
PROPOSITION XXIII. PROBLEM.
At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.
Let AB be the given st. line, and A the given pt. in it, and DCE the given rectil. L.
It is required to make an at A, in the given line AB, that shall be = the given rectil. _ DCE.
Constr. In CD, CE take any pts. D, E, and join DE;
make A AFG whose sides shall = the three st. lines CD, DE, EC, so that AF=CD, AG=CE, FG=DE.
(i. 22) Then shall Z FAG= _ DCE. Dem. :: FA, AG= DC, CE, each to each, and base FG= base DE;
(constr.) .. L FAG= L DCE.
(i. 8) Wherefore, at the given pt. A in the given st. line AB, FAG is made = L DCE.
Q. E. F.
PROPOSITION XXIV. THEOREM.
If two triangles have two sides of the one equal to two sides of the other, each to each, but the angie contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the other.
Let ABC, DEF be two As, which have AB, AC =DE, DF, each to each, namely AB = DE, and AC = DF; but 2 BAC greater than _ EDF.
Then shall base BC be greater than base EF.
Constr. Of the two sides DE, DF, let DE be not greater than DF; at D in DE, make 4 EDG= L BAC;
(i. 23.) make DG = DF or AC, (i. 3) and join EG, GF. Dem. Then :::DE = AB, and DG AC, (hyp.) the two sides DE, DG = AB, AC each to each, and / EDG < BAC; ... base EG = base BC.
(i. 4) And :: DG DF, LDFG = L DGF; (i. 5) but _ DGF is greater than _ EGF; (ax. 9)
2 DFG is greater than L EGF; .:. much more is _ EFG greater than 6 EGF. And .. in A EFG, L EFG is greater than _ EGF,
..EG is greater than EF; (i. 19)
But EG was proved = BC; .. BC is greater that EF. Wherefore, if two triangles, &c.
Q. E. D.
[Do not forget to insert the restriction of the two sides DE, DF, let DE be not greater than DF. Without this there would be three cases, for F might fall on, above, or below EG. But with this condition F must fall below EG, for if DF cuts EG in H, 4 DHG is greater than 2 DEG,
(i. 16) and < DEG is not less than 2 DGE, since by supposition
DF (which DG) is not less than DE, (i. 18)
(i. 19) or, is less than DF, since DF = DG.]
PROPOSITION XXV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other; the angle contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides, equal to them, of the other.
Let ABC, DEF be two As which have AB, AC equal to DE, DF, each to each, namely AB = DE, and AC = DF; but base BC greater than base EF.
Then shall Z BAC be greater than EDF.
Dem. For, if _ BAC be not greater than 2 EDF, it
must either be equal to it, or less than it. If L BAC = L EDF, then base BC = base EF; (i. 4) but it is not equal, .. _ BAC is not LEDF.
Again, if L BAC were less than 2 EDF,
base BC would be less than base EF; (i. 24) but it is not less, LBAC is not less than _ EDF; and it has been shown, that BAC is not = L EDF;
BAC is greater than < EDF.