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Produce BA to the point D, and make (I. 3.) AD equal to AC; and join DC. Because DA is equal to AC, likewise (I. 5.)

1. The angle ADC is equal to ACD; but the angle BCD is greater than the angle ACD; therefore

2. The angle BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater (I. 19.) side is opposite to the greater angle; therefore

3. The side DB is greater than the side BC; but DB is equal to BA and AC; therefore

4. The sides BA, AC, are greater than BC. In the same manner it may be demonstrated that

5. The sides AB, BC, are greater than CA; and BC, CA, are greater than AB. Therefore any two sides, &c. Q.E.D.

PROP. XXI.-THEOREM.

If from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let the two straight lines BD, CD, be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA, AC, of the triangle, but contain an angle BDC greater than the angle BAC.

E

B

Produce BD to E; and because two sides of a triangle are greater than the third side (I. 20.)

1. The two sides BA, AE, of the triangle ABE, are greater than BE; to each of these add EC; therefore

2. The sides BA, AC, are greater than BE, EC. Again, because (I. 20.)

3. The two sides CE, ED, of the triangle CED, are greater than CD,

add DB to each of these; therefore (Ax. 4.)

4. The sides CE, EB, are greater than CD, DB; but it has been shown that BA, AC, are greater than BE, EC; much more then

5. BA, AC, are greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle (I. 16.)

1. The exterior angle BDC of the triangle CDE is greater than CED. For the same reason,

2. The exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC' is greater than CEB; much more then

3. The angle BDC is greater than the angle BAC. Therefore, if from the ends of, &c. Q.E.D.

PROP. XXII. - PROBLEM.

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, (I. 20.)

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 Č; A and C greater than B; and B and C than 4. It is required to make a triangle, of which the sides shall be equal to A, B, C, each to each.

Take a straight line DE terminated at the point D, but unlimited towards E, and make (I. 3.) DF equal to 1, FG equal to B, and GH equal to C'; and from the centre F, at a distance FD, describe (Post. 3.) the circle DKL; and from the centre G, at the distance GH, describe another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines A, B, C.

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Because the point F is the centre of the circle DKL, (Def. 15.)

1. FD is equal to FK; but FD is equal to the straight line 4; therefore

2. FK is equal to A.

Again, because G is the centre of the circle LKH,

3. GH is equal to GK; but GH is equal to C; therefore also

4. GK is equal to C; and FG is equal (Constr.) to B; therefore the three straight lines KF, FG, GK, are equal to the three A, B, C. And therefore

5. The triangle KFG has its three sides KF, FG, GK, equal to the three given straight lines A, B, C. Which was to be done.

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

Take in CD, CE, any points D, E, and join DE; and make (I. 22.) the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, CE, so that CD be equal to AF, CE to AG, and DE to FG; the angle FAG is equal to the angle DCE.

A

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Because DC, CE, are equal to FA, AG, each to each, and the base DE to the base FG; (I. 8.)

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.-THEOREM.

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 is also greater than the base EF.

D

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Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make (I. 23.) the angle EDG equal to the angle BAC; and make DG equal (I. 3.) to AC or DF, and join EG, GF.

Because AB is equal to DE, and AC to DG, the two sides BA, AC, are equal to the two ED, DG, each to each, and the angle BAC is equal (Constr.) to the angle EDG; therefore (I. 4.)

1. The base BC is equal to the base EG. And because DG is equal to DF, (1, 5.)

2. The angle DFG is equal to the angle DGF; but the angle DGF is greater than the angle EGF, therefore

3. The angle DFG is greater than EGF; and much more

4. The angle EFG is greater than the angle EGF. And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater (I. 19.) side is opposite to the greater angle; therefore

5. The side EG is greater than the side EF; but EG is equal to BC; and therefore also

6. BC is greater than EF. Therefore, if two triangles, &c. Q.E.D.

PROP. 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 the one greater than the base of the other; the angle also 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. AB equal to DE, and AC to DF; but the base CB greater than the base EF; the angle BAC is likewise greater than the angle EDF. A

D

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For if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle EDF, because then the base BC would be equal (I. 4.) to EF; but it is not, therefore

1. The angle BAC is not equal to the angle EDF. Neither is it less; because then the base BC would be less (I. 24.) than the base EF; but it is not, therefore

2. The angle BAC is not less than the angle EDF; and it was shown that it is not equal to it; therefore

3. The angle BAC is greater than the angle EDF. Wherefore if two triangles, &c. Q.E.D.

PROP. XXVI.-THEOREM.

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.

Let ABC, DEF, be two triangles which have the angles ABC, BCA, equal to the angles DEF, EFD, 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 are equal in the two triangles, viz. BC to 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.

D

A

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For if AB be not equal to DE, one of them must be the greater. 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,

1. The two sides GB, BC, are equal to the two DE, EF, each to each; and the angle GBC' is equal to the angle DEF; therefore (I. 4.)

2. 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

3. The angle GCB is equal to the angle DFE; but DFE is, by the hypothesis, equal to the angle BCA; wherefore also

4. The angle BCG is equal to the angle BCA,

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