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Hypothesis.-When two triangles have two angles of the one respectively equal to two angles of the other and the side adjacent to the two angles of the one equal to the side adjacent to the two angles of the other,

Conclusion-The third angle of the one is equal to the third angle of the other, and the remaining sides of the one are respectively equal to the remaining sides of the other.

Proofs of Equality.-To prove the equality of two triangles, we place one on the other, and show that they coincide.

For example, let us suppose that AB and D E are two sides which we know to be equal. We arrange the superposition by three steps:1. Place ▲ DEF on A ABC (Fig. 51a),

2. So that the point D is on the point A (Fig. 516),

3. And so that the line D E is on the line A B (Fig. 51c),

FIG. 510

B

D.A

FIG. 516.

FIG. 51C.

The triangles are now completely superposed. It would be wrong to add, "and so that DF lies on A C;" for this follows as a natural consequence of the previous superposition. When we have fixed one point and a line proceeding from that point, any subsequent statement of coincidences must commence with such words as properly introduce a consequence; thus we can say then, since the angle D is equal to the angle A, the side D F will fall on A C."

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Easy Exercises.

(a). Make a right angle with your set-square; and then draw a circle, so that the centre shall be the vertex of the angle; and then say how much of the circle is within the angle.

(b). Draw a circle, having a radius of one inch.

(c). Draw two circles, each having a radius of one inch and a half.

(d). Draw a straight line, A B, one inch and a half long.

D

THEOREMS ON CHAPTER IV.

SECTION I.

DEFINITIONS.

31. A triangle is a closed figure bounded by three straight lines which meet in three points and are called the sides of the triangle.

THEOREM V.

When two triangles have two sides of the one respectively equal to two sides of the other, and the angles included by them equal, then shall the triangles be equal in all respects; thus the third side of the one shall be equal to the third side of the other, and those angles shall be equal which are opposite to the equal sides.

AA

Let ABC and DEF be two triangles, having

AB DE, AC=DF, and BAC = 4 ED F.

Then shall ▲ ABC be equal to ▲ DEF in all respects.

Suppose A ABC applied to A DEF, so that the

point A may fall on D,

and the line A B along D E.

Then because A B D E the point B will fall on E.

=

And because ▲ BAC = 4 EDF the line A C will fall along D F.

And because ACDF the point C will fall on F.

Hence BC coincides with EF, and ▲ ABC coincides with A DEF, and is equal to it in all respects. Consequently BC EF, 4 ABC = 4 DEF, and LACB=4DFE.

=

THEOREM VI.

When two triangles have two angles of the one respectively equal to two angles of the other, and the side common to the two angles of the one equal to the side common to the two angles equal to them of the other, the triangles are equal in all respects, and those sides are equal which are opposite to the equal angles.

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Suppose A ABC placed on A DEF, so that the point B may fall on E, and BA along ED; then, because ▲ A B C = ▲ D E F by hypothesis, B C will fall along E F.

Because B C E F by hypothesis, the point C will fall on F.

Because

BCA EFD by hypothesis, CA will

fall along F D.

=

Hence the point A must fall on D, and ▲ A B C must coincide with A DEF and be equal to it in all respects.

THEOREM VII.

The exterior angle of a triangle is greater than either of the interior and non-adjacent angles.

Hypothesis. Let ACD be the exterior angle of A A B C, then. Conclusion ACD is greater CA B or 4 CBA.

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drawn and produced to E, so that FE = F B, and the points C and E joined by the straight line CE.

=

Then in the triangles BAF, ECF, FC = FA, FE FB, and ▲ BFA = EFC; therefore the triangles are equal in all respects, and FCE = < FAB.

Consequently 4 A.C D, which is greater than ▲ FCE, must also be greater than ▲ FAB.

By continuing the side A C, it may be proved in a similar manner that the angle vertically opposite to ACD is greater than ▲ CBA. Therefore also is greater than ▲ CBA.

THEOREMS FOR EXERCISE.

ACD

6. Two straight lines drawn from any point in the perpendicular bisector of a straight line to the extremities of the straight line are equal to one another.

7. If from a point within a triangle two straight lines. be drawn to the extremities of the base, the angle they contain shall be greater than the vertical angle of the triangle.

8. Only one perpendicular can be drawn from a point to a straight line.

SECTION II.-Relations between the Sides and Angles of a Triangle.

We proceed now to consider the relations of equality and inequality that exist between the sides and angles of a triangle.

The Isosceles Triangle.—A triangle like that represented in Fig. 52, having two sides, A B and A C, equal to one another, is termed isosceles. It will be shown that such a triangle possesses two important properties: (a) it is symmetrical about the straight line which bisects the vertical angle; (b) it is reversible.

B

D

FIG. 52.

Propositions on the Isosceles Triangle.-From the same point draw two straight lines so that they form an angle, A. On one of the sides of this angle mark off any length, AB, and on the other side mark off a length, A C, equal to A B. Now draw the line, B C. The figure ABC is an isosceles triangle, for it has two sides, A B and A C, equal to one another. In constructing the triangle, we paid no regard to the angles B and C at the base of the triangle; but when the triangle is made, it is found that these angles are equal to one another.

The proposition here suggested may be enunciated as follows :— Hypothesis.—Let two sides of a triangle be equal to one another; Conclusion-Then shall the angles opposite these sides be equal to one another.

It is generally very easy to point out which is the hypothesis and which is the conclusion, but sometimes this is not the case. For instance, the above proposition is frequently stated thus: In an isosceles triangle the angles at the base are equal. Here the hypothesis is expressed in the word isosceles, which means having two sides equal.

Let us take another illustration :-Draw a straight line, B C, and at B and C let two acute angles, CBA, BCA, be made equal to one another. Let CA and B A meet in the point A, then the triangle A B C will have been constructed without any reference to the magnitude of the sides CA and BA, but it will be found that these sides are equal. The enunciation of the proposition here indicated will therefore be :

Hypothesis. When two angles of a triangle are equal,
Conclusion-The sides opposite to them are equal.

B

FIG. 53.

c

Compare carefully the statements of these two propositions. We

will repeat them.

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