PROP. XXII. PROB.

To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third.* Let A, B, C, be the three given straight lines, of which any two whatever are greater than the third, viz. A and B greater than C; A and C greater than B; and B and C greater than A. It is required to make a triangle of which the sides shall be equal to ABC, each to each. Take a straight line DE terminated at the point D, but unlimited

20.1.

* 3 Post.

DKL; and from the centre G, at the distance
GH, describe* another circle HLK; and join *3 Post.
KF, KG: the triangle KFG shall have its sides
equal to the three straight lines A, B, C.

+1. Ax.

Because the point F is the centre of the circle DKL, FD is equal* to FK; but FD is equal to *15 Def. the straight line A; therefore FK is equal to A: Const. Again, because G is the centre of the circle LKH, GH is equal to GK; but GH is equal to *15 Def. C; therefore also GK is equal to C: and FG is equal to B: therefore the three straight lines +Const. KF, FG, GK, are equal to the three A, B, C: and therefore the triangle KFG has its three sides KF, FG, GK, equal to the three given straight lines Á, B, C. Which was to be done.

PROP. XXIII. PROB.

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 straight line, and A the given point in it, and DCE the given rectilineal

D

angle; it is required to make an angle at the
given point A in the
given straight line AB,
that shall be equal to
the given
angle DCE.

rectilineal

AA

În CD, CE, take any D4 points D,E, and join DE;

22. 1. and make* the triangle

*8.1.

B

G

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.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle 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 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. Because AB is equal to DE, and AC† to DG, Const. the two sides, BA, AC, are equal to the two ED,

3.1.

+Hyp.

*

DFG is equal to the angle DGF; but the angle *E. 1. DGF is greater 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.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the 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 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 be equal to EF: but it

В

D

*4. 1.

*

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 not; therefore the angle BAC is not less than the angle EDF: and it was shown that it is not equal to it; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E. D.

+3.1.

+ Hyp.

PROP. XXVI. THEOR.

If two triangles have two angles of the one equal to two angles of the other, each to each; and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each; then shall the other sides be equal, each to each, and also the third angle of the one to the third angle of the other.

Ꭰ

Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, each to each, viz. ABC to ĎEF, 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 A are equal in the two triangles, viz. BC to G EF: the other sides shall be equal, each to each, viz. AB to DE, and AC to DF, and the third angle BAC to the third angle EDF

B

CE

For, if AB be not equal to DE, one of them must be greater than the other. Let AB be the greater of the two, and make BG equal† to DE, and join GC: therefore, because BG is equal to 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; there4. 1. fore the base GC 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 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 equal to † Hyp. EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle, ABC is equal 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 which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE: likewise in this 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 BHI equal† to EF, and †3. 1. join AH and because BH is equal to EF, and AB tot 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† +Hyp. to the base DF, and the triangle ABH to the +41 triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite: therefore the angle BHA is equal to the angle EFD: but EFD is equal to the angle † Hyp. BCA; therefore also the angle BHA is equal to †1 Ax. 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 equal to DE; therefore the + Hyp. two AB, BC are equal to the two DE, EF, each to each; and they contain † equal angles; where- +Hyp. fore the base AC is equal† to the base DF, and †4. 1.