LIMITS OF SIDES.

266. Theorem.-Each side of a triangle is smaller than the sum of the other two, and greater than their difference.

The first part of this theorem is an immediate consequence of the Axiom of Distance

(54); that is,

AC <AB+ BC.

Subtract AB from both members of this inequality, and AC AB BC.

That is, BC is greater than the difference of the other sides.

Prove the same for each of the other sides.

267. An EQUILATERAL triangle is one which has three sides equal.

An ISOSCELES triangle is one which has only two sides equal.

A SCALENE triangle is one which has no two sides equal.

EQUAL SIDES.

268. Theorem.-When two sides of a triangle are equal, the angles opposite to them are equal.

If the triangle BCD is isosceles, the angles B and D, which are opposite the equal sides,

are equal.

Let the angle C be divided into two equal parts, and let the dividing line extend to the opposite side of the triangle at F.

B

Then, that portion of the figure upon one side of this line may be turned

Geom.-8

upon an axis. Since the angle C was bisected, the line BC will fall upon DC; and, since these two lines are equal, the point B will fall upon D. But F, being a point of the axis, remains fixed; hence, BF and DF will coincide. Therefore, the angles B and D coincide, and are equal.

269. Corollary. The three angles of an equilateral triangle are equal.

270. In an isosceles triangle, the angle included by the equal sides is usually called the vertex of the triangle, and the side opposite to it the base.

271. Corollary.—If a line pass through the vertex of an isosceles triangle, and also through the middle of the base, it will bisect the angle at the vertex, and be perpendicular to the base.

The straight line which has any two of these four conditions must have the other two (52).

UNEQUAL SIDES.

272. Theorem. -When two sides of a triangle are unequal, the angle opposite to the greater side is greater than the angle opposite to the less side.

If in the triangle BCD the side BC is greater than DC, then the angle D is greater

than the angle B.

Let the line CF bisect the angle C, and be produced to the side BD. Then let the triangle CDF turn upon CF. CD will take the direction CB; but, since CD is less than CB, the point D will fall between C and B, at G. Join GF.

B4

F

Now, the angle FGC is equal to the angle D, because

they coincide; and it is greater than the angle B, because it is exterior to the triangle BGF (261). Therefore, the angle D is greater than B.

273. Corollary.-When one side of a triangle is not the largest, the angle which is opposite to that side is acute (257).

274. Corollary.—In a scalene triangle, no two angles are equal.

EQUAL ANGLES.

275. Theorem.-If two angles of a triangle are equal, the sides opposite them are equal.

For if these sides were unequal, the angles opposite to them would be unequal (272), which is contrary to the hypothesis.

276. Corollary.—If a triangle is equiangular, that is, has all its angles equal, then it is equilateral.

UNEQUAL ANGLES.

277. Theorem.-If two angles of a triangle are unequal, the side opposite to the greater angle is greater than the side opposite to the less.

If, in the triangle ABC, the angle C is greater than the angle A, then AB is

greater than BC.

For, if AB were not greater

than BC, it would be either

B

equal to it or less. If AB were equal to BC, the opposite angles A and C would be equal (268); and if AB were less than BC, then the angle C would be less than A (272); but both of these conclusions are contrary to the hypothesis. Therefore, AB being neither less than nor equal to BC, must be greater.

278. Corollary.—In an obtuse angled triangle, the longest side is opposite the obtuse angle; and in a right angled triangle, the longest side is opposite the right angle.

279. The HYPOTENUSE of a right angled triangle is the side opposite the right angle. The other two sides are called the legs.

The student will notice that some of the above propositions are but different statements of the principles of perpendicular and oblique lines.

EXERCISES.

280.-1. How many degrees are there in an angle of an equilateral triangle?

2. If one of the angles at the base of an isosceles triangle be double the angle at the vertex, how many degrees in each?

3. If the angle at the vertex of an isosceles triangle be double one of the angles at the base, what is the angle at the vertex? 4. To circumscribe a circle about a given triangle (149). 5. To inscribe a circle in a given triangle (252).

6. If two sides of a triangle be produced, the lines which bisect the two exterior angles and the third interior angle all meet in one point.

7. Draw a line DE parallel to the base BC of a triangle ABC, so that DE shall be equal to the sum of BD and CE.

8. Can a triangular field have one side 436 yards, the second 547 yards, and the third 984 yards long?

9. The angle at the base of an isosceles triangle being onefourth of the angle at the vertex, if a perpendicular be erected to the base at its extreme point, and this perpendicular meet the opposite side of the triangle produced, then the part produced, the remaining side, and the perpendicular form an equilateral triangle.

.10. If with the vertex of an isosceles triangle as a center, a circumference be drawn cutting the base or the base produced, then the parts intercepted between the curve and the extremities of the base, are equal.

EQUALITY OF TRIANGLES.

281. The three sides and three angles of a triangle may be called its six elements. It may be shown. that three of these are always necessary, and they are generally enough, to determine the triangle.

THREE SIDES EQUA L.

282. Theorem.-Two triangles are equal when the three sides of the one are respectively equal to the three sides of the other.

Let the side BD be equal to AI, the side BC equal to AE, and CD to EI; then the two triangles are equal.

A

I

Now, the point A being on B, the points E and C are at the same distance from B, and therefore they are both in the circumference, which has B for its center, and BC or AE for its radius (153). For a similar reason, the points E and C are both in the circumference, which has D for its center and DC or IE for its radius. These two circumferences have only one point common on one side of the line BD, which joins their centers (232). Hence. E and C are both at this point. Therefore (51), AE coincides.