Then ABC, A'C'B' will be two triangles, having the sides A B, A C, equal to the sides A'C', A'B', respectively, and the angle BAC equal to C'A'B'. Therefore the remaining angles of the one are equal to the remaining angles of the other; that is to say, the angle A B C is equal to A'C'B', and therefore to the angle AC B.

Next, let the angle ABC equal AC B. The side A B shall be equal to A C.

The same construction being made, the triangles ABC, A'C' B', will be triangles having the angles A B C, A CB, respectively equal to angles A'C'B', A'B'C', and the side B C equal to C'B'. Therefore the remaining sides are equal, and AB is equal to A'C' or to A C.

Cor. If the angles at the base of a triangle are not equal, the opposite sides are unequal, and if the sides of a triangle are not equal the opposite angles are unequal.

XIX.-EUCLID I. 18.

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

Let A B C (Fig. 13) be a triangle of which the side A C is greater than A B; the angle ABC shall be greater than A C B.

Because A C is greater than A B, let A D, part of A C, be equal to A B; join BD. Then because

AD B is the exterior angle of the triangle BDC, it is greater than the interior and opposite angle DC B. But ADB is equal to ABD, because A B D is an isosceles triangle. Therefore the angle ABD is likewise greater than the angle AC B. Much more then is the angle ABC (which consists of ABD and D B C) greater than AC B.

XX.

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

Let ABC be a triangle, of which the angle A B C is greater than ACB; the side A C will be greater than A B.

Because the angles at the base of the triangle A B C are not equal, the opposite sides are unequal (Prop. 18, Cor.); that is, one of the two A B, A C, is greater than the other. But in a triangle having one side greater than another, the angle opposite the greater side is greater than the angle opposite the less; and in the triangle ABC the angle ABC is greater than A CB; therefore the angle A B C is opposite the greater of the two A B, A C, or, in other words, A C is greater than A B.

XXI.-EUCLID I. 20.

Any two sides of a triangle are together greater

than the third side.

Let A B C be a triangle (Fig. 14), take any side BA and produce it at one extremity to D, making A D equal to AC the adjacent side, and join D C.

Because AD is equal to A C, the angle ADC is equal to A CD, and the angle B C D is equal to BCA and A CD, and is therefore greater than ACD, or ADC; and because the angle BDC is a triangle in which the angle B C D is greater than BDC, therefore the side DB opposite BCD is greater than the side B C opposite BDC. But DB is equal to BA and AD; that is, to B A and AC. Therefore B A and A C, or any two adjacent sides, are greater than the third side B C.

XXII.-EUCLID I. 21.

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

Let the two straight lines B D, CD (Fig. 15) be drawn from B, C, the ends of the side B C of the triangle ABC, to a point D within it; BD and CD together shall be less than the other sides BA, AC of the triangle, but shall contain an angle B DC greater than the angle B A C.

Produce B D to meet the side of the triangle in E. Then the two sides BA, AE of the triangle ABE are greater than the third side B E. To each of these add E C. Therefore the sides B A, A C are

greater than BE, E C. Again, because the two sides CE, ED of the triangle CED are greater than CD, to each of these add DB; therefore the sides CE, E B are greater than CD, D B. Much more then are the sides BA, AC greater than BD, DC.

Again, because the exterior angle of a triangle is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for a like reason, the angle BEC is greater than BA C. Much more then is the angle BDC greater than BA C.

XXIII.-EUCLID I. 24.

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 the one 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 (Fig. 16) be two triangles, having the two sides A B, AC equal to the two DE, DF each to each; and the angle BAC greater than the angle EDF; the base B C shall also be greater than the base E F.

Of the sides DE, DF let D E be the side which is not greater than the other, and let D G be a straight line making the angle ED G equal to

BAC; let D G be equal to AC or DF; GF, GE straight lines joining G with F and E.

Then, because A B is equal to DE, and A C to D G, and the angle BAC to ED G, therefore the base BC is equal to E G; and because DG is equal to D F, the angle DFG is equal to DG F, but the angle DG F is greater than the angle EG F, and much more is the angle EF G greater than the angle E G F; therefore the side EG of the triangle E G F, which is opposite the angle EFG, is greater than E F, the side opposite EFG; but EG is equal to B C, therefore B C is greater than E F.

XXIV.

If two triangles have two sides of the one respectively equal to two sides of the other, 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 is greater than the angle contained by the sides equal to them of the other triangle.

In the triangles A B C, D E F (Fig. 17) let A B, AC be equal to DE, DF respectively, and let EF be greater than B C. Then the angle EDF shall be greater than the angle BA C.

Because the bases BC, EF are not equal (by Prop. 16, Cor.), the opposite angles BA C, EDF are unequal, or one of them is the greater of the