PROP. XVIII. THEOR.

If one side of a triangle be greater than another, the angle opposite to the greater side, is greater than the angle opposite to the less.

In the triangle ABC, let the side AB be greater than the side BC; then the angle BCA is greater than the angle BAC.

A

D

B

C

From the greater side AB, cut off a part BD, equal to the less BC (Prop.3.); join DC. Because the sides BD and BC are equal (Const.), the angles BCD and BDC are eqnal (Prop.5.); but the angle BDC is greater than the interior angle BAC (Prop. 16.), therefore the angle BCD is greater than BAC; much more then is the angle B C A greater than the angle BA C. Which was to be demonstrated.

PROP. XIX. THEOR.

If one angle of a triangle be greater than another, the side opposite to the greater angle, is greater than the side opposite to the less.

In the triangle ABC, let the angle c be greater than the angle A; then the side AB is greater than the side BC.

For the side AB is either equal to BC,

C

B

or less than it, or greater than it. AB is not equal to в C, for, if it were, the angles at a and c would then be equal (Prop. 5.); but they are not (Hyp.). Again, AB is not less than BC, for, if it were, the angle c would then be less than the angle A (Prop. 18.); but it is not (Hyp.): since then AB is neither equal to, nor less than BC, it must be greater than it. Which was to be demonstrated

SCHOL. One side of a triangle is greater than, equal to, or less than another, according as the angle opposite to the side is greater than, equal to, or less than the angle opposite to the other.

PROP. XX. THEOR.

Any two sides of a triangle are together greater than the third

side.

In the triangle ABC, the sides AB and BC are greater than AC, the sides BC and CA are greater than BA, and the sides CA and A B are greater than BC.

Bisect the angle ABC by the line BD (Prop. 9.).

Then the angle ADB is greater than the A interior angle DBC, also the angle CDB is

greater than the interior angle DBA (Prop. 16.); but the angles DBA and DBC are equal (Const.); therefore the angle ADB is greater than DBA, and the angle CDB is greater than DBC, and therefore the sides AB and BC opposite the greater angles, are greater than the sides AD and DC opposite to the less (Prop. 19.); therefore the two sides A B and BC are greater than the third side A c. In like manner, by bisecting any other angle, it can be shown that the sides containing it are greater than the third side. W. W.D.

Two right lines drawn from the ends of one side of a triangle to a point within it, are less than the two other sides of the triangle; but contain a greater angle.

In the triangle ABC, the right lines AD and CD, drawn from a and c to the point D within the triangle, are less than the sides AB and BC; but contain an angle ADC greater than ABC.

B

E

C

Produce AD to E; then in the triangle ABE, the sides AB and BE are greater than the third side AE (Prop. 20.); add EC to each, then AB and BC are greater than AE and EC; but in the triangle CED, the sides DE and EC are greater than the third side DC (Prop. 20.); add AD to each; then AE and EC are greater than AD and DC; much more then are the sides AB and BC greater than AD and DC. Again, the exterior angle AEC is greater than the interior angle ABC (Prop. 16.), and the exterior angle ADC is greater than AEC (Prop. 16.); therefore the angle ADC is greater than Which was to be demonstrated.

ABC.

PROP. XXII. PROB.

To construct a triangle which shall have its sides respectively equal to three given lines, of which any two are greater than the third.

with the radius DF describe a circle, and from the centre E with the radius EG describe another circle, cutting the former in H; join DH and EH; the sides of the triangle DEH are equal to the lines A, B, C. For the line DH is equal to DF, being radii of the same circle (Def. 16.); but DF is equal to B (Const.), therefore DH is equal to в (Ax. 1.). Also the line EH is equal to EG, being radii of the same circle (Def. 16.); but EG is equal to c (Const.), therefore E H is equal to c; and DE is equal to a (Const.); therefore the three sides DH, EH, and DE of the triangle DE H are equal to the given lines A, B, and c. Which was to be done.

SCHOL. Another triangle having its sides equal to the three given lines, may be obtained by drawing right lines from D and E to the point L, in which the circles cut at the other side of DE.

PROP. XXIII. PROB.

At a point in a given line, to make an angle equal to a given one.

Let A be the given point, AB the given line, and D the given angle; it is required to make an angle at a equal to the angle D,

In the sides of the angle D, take any two points c and E; join CE, and construct a triangle BAF, having its

D

A

E

C

F

B

sides A B, BF, and FA equal to the lines DC, CE, and ED (Prop. 22.); then the angle a is equal to the angle D (Prop. 8). W. W. D.

If two triangles have two sides of the one, respectively equal to two sides of the other, and if the contained angle in the one be greater than the contained angle in the other; the base opposite to the greater angle is greater than the base opposite to the less.

In the triangles ABC and DEF, let the side A B be equal to DE, the side BC to EF: but the angle DEF greater than the angle ABC; then the base DF is greater than AC.

At the point E with the less side DE, make an angle DEG equal to the angle ABC (Prop.

ДД

23.), and make EG equal to B C or E F (Prop. 3.); join G F and GD. Then in the triangles A B C and DEG, the sides AB and BC in the one, are equal to the sides DE and EG in the other (Const.), and the angle B is equal to the angle DEG (Const.); therefore the base D G is equal to AC (Prop. 4.). And as the sides EF and EG are equal (Const.), the angles EGF and EFG are equal (Prop. 5.); but the angle EFG is greater than DFG, therefore the angle EGF is greater than D F G ; much more then is the angle DGF greater than DFG: therefore in the triangle DFG, the side D F is greater than DG (Prop. 19.); but DG is equal to a c, therefore D F is greater than a c. W. W. D.

PROP. XXV. THEOR.

If two triangles have two sides of the one respectively equal to two sides of the other, and if the base of one be greater than the base of the other; the angle opposite to the greater base is greater than the angle opposite to the less.

In the triangles A B C and D E F, let the side A B be equal to D E, and BC to EF; but the base D F greater than AC the angle E is greater than the angle B.

A

B

C D

E

F

For the angle E is either less than, equal to, or greater than the angle B. It is not less than it, because the base DF would then be less than AC (Prop. 24.), but it is not (Hyp.). It is not equal to it, because the bases DF and AC would then be equal (Prop. 4.), but they are not (Hyp.): since then the angle E is neither less than, nor equal to the angle B, it must be greater than it. W. W. D.

PROP. XXVI. THEOR.

If two triangles have two angles and a side in the one, respectively equal to two angles and a side similarly situated in the other, the remaining sides and angle in the one, are respectively equal to the remaining sides and angle in the other, and the two triangles are also equal.

In the triangles A B C and D E F, let the angle BAC be equal to EDF, the angle A C B to D F E, and a side in one triangle, equal to a side similarly situated in the other; the remaining sides and angles are respectively equal.

G

B

E

с

D

CASE 1. When the equal sides A are A C and D F adjacent to the equal angles, then the side A B is equal to D E, and the triangles are also equal.

For, if AB and DE be not equal, let one of them A B be greater than the other, and from it cut off AG equal to D E, and join CG. Then the sides AC and AG are equal to the sides D F and DE, and the angles at A and D are equal (Hyp.); therefore the angle ACG is equal to DFE (Prop. 4.): but the angle ACB is also equal to DFE (Hyp.); therefore A C G is equal to A CB, a part equal to the whole, which is absurd; therefore A B and DE are not unequal, therefore they are equal; and also A c is equal to DF, and the angles at A and D are equal (Hyp.); therefore B C is equal to E F, the angle в to the angle E, and the triangles A B C and D E F are equal in all respects (Prop. 4.). CASE 2. When the equal sides are AB and DE opposite to the equal angles, then A C is equal to D F.

A

B

E

G

C D

P

For, if AC and D F be not equal, let one of them A c be greater than the other, and from it cut off AG equal to DF, and join BG. Then the sides AG and AB are equal to the sides DF and DE, and the angles at A and D are equal (Hyp.); therefore the angle A G B is equal to DFE (Prop. 4.); but the angle ACB is also equal to DFE (Hyp.), therefore AGB is equal to a C B, an exterior angle equal to its interior opposite angle, which is impossible (Prop. 16.); therefore AC and Dr are not unequal, therefore they are equal; and also A B is equal to DE, and the angles at A and D are equal (Hyp.); therefore BC is equal to EF, the angle в to the angle E, and the triangles A B C and D E F are equal in all respects (Prop. 4.). W. W.D.