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* 3 Posts
PROP. XXII. PROB.
to three given straight lines, but any two what-
Let A, B, C, be the three given straight lines,
Take a straight line DE terminated at the
Because the point F is the centre of the circle DKL, FD is equal* to FK; but FD is equal t to * 15 Def. the straight line A; therefore FK is equal tto A: Const
PROP. XXIII. PROB.
make a rectilineal angle equal to a given recti-
Let AB be the given straight line, and A the
angle; it is required to make an angle at the
B * 22. 1. and make* the triangle
AFG, the sides of which shall be equal to the three straight lines CD, DE, EC, so that CD be equal to AF, CE to AG, and DE to FG: the angle FAG shall be equal to the angle DCE.
Because DC, CE are equal to FA, AG, each to each, and the base DE to the base FG: the angle DCE is equal * to the angle FAG. Therefore at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done.
PROP. XXIV. THEOR.
to two sides of the other, each to each, but the
Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF: the base BC shall be greater than the base EF.
Of the two sides DE, DF, let DE be the side
which is not greater than the other, and at the *23. 1. point D, in the straight line DE, make* the
angle EDG equal to the angle BAC; and make
DG equal* to AC or DF, and join EG, GF. + Hyp. Because AB is equal to DE, and AC+ to DG, + Const the two sides, BA, AC, are equal to the two ED,
DG, each to each,
D and the angle BAC is equalt to the angle
+ Const. EDG; therefore the hase BC is equal* to the base EG. And
B because DG is equal to DF, the angle DFG is equal* to the angle DGF; but the angle * 5. 1. DGF is greatert than the angle EGF; therefore +9 ax, the angle DFG is greater than EGF; therefore much more is the angle EFG greater than the angle EGF: and because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater* angle is subtended by the *19.1. greater side, therefore the side EG is greater than the side EF: but EG was proved to be equal to BC; therefore BC is greater than EF. Therefore if two triangles, &c. Q. E. D.
PROP. XXV. THEOR.
to two sides of the other, each to each, but the
Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. A3 equal to DE, and AC to DF; but the base BC greater than the base EF: the angle BAC shall be greater than the angle EDF.
For, if it be not greater, it must either be equal to it, or less than it: but the angle BAC is not equal to the angle EDF, because then the base BC would
B be equal* to EF: but it
+ Hyp. ist not: therefore the angle BAC is not equal to
the angle EDF: neither is it less, because then * 24. 1. the base BC would be less * than the base EF: + Hyp. but it is t not; therefore the angle BAC is not
less than the angle EDF: and it was shown that
PROP. XXVI. THEOR.
to two ungles of the other, each to each ; and
Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, each to each, viz. ABC to DEF, and BCA to EFD); also one side equal to one side; and first let those sides be equal which are adjacent to the angles that
For, if AB be not equal to DE, one of them
must be greater than the other. Let AB be the + 3. 1. greater of the two, and make BG equal t to DE,
and join GC: therefore, because BG is equal to + Hyp. DE, and BC+ to EF, the two sides GB, BC are
equal to the two DE, EF, each to each ; and the Hyp. angle GBC is equal to the angle DEF; there
fore the base G is equal* to the base DF, and the triangle GBC to the triangle DEF, and the other angles to the other angles, each to each, to
which the equal sides are opposite : therefore the angle GCB is equal to the angle DFE: but DFE is, by the hypothesis, equal to the angle BCA; wherefore also the angle BCG is equal.t to the +1 Ax. angle BCA, the less to the greater, which is impossible: therefore AB is not unequal to DE, that is, it is equal to it: and BC is equalt to + Hyp. EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle, ABC is equalt to the angle DEF; therefore the base + Hyp. AC is equal * to the base DF, and the third angle *4. 1. BAC to the third angle EDF. Next, let the sides
D which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE: likewise in this
B HC E case, the other sides shall be equal, AC to DF, and BC to EF; and also the third angle BAC to the third angle EDF.
For, if BC be not equal to EF, let BC be the greater of them, and make BII equalt to EF, and +3. 1. join AH : and because BH is equal to EF, and AB to + DE; the two AB, BH are equal to the + Hyp. two DE, EF, each to each ; and they contain equal angles ; therefore the base AH is equal t *Hyp. to the base DF, and the triangle ABH to the + 4.1 triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite : therefore the angle BHÁ is equal to the angle EFD: but EFD is equalt to the angle + Hyp. BCA,
therefore also the angle BHA is equalt to †1 Aä. the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA; which is impossible : * *16. 1. wherefore BC is not unequal to EF, that is, it is equal to it: and AB is equalf to DE; therefore the f Hyp. two AB, BC are equal to the two DE, EF, each to
and they contain t equal angles; where- + Hyp. fore the base AC is equalt to the base DF, and +4. I.