Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

EXERCISES ON PROPOSITIONS 1 TO 3.

a

a

1. If the two circles in Proposition 1 cut one another again at F, prove that AFB is an equilateral triangle.

2. If the two circles in Proposition 1 cut one another at C and F, prove that the figure ACBF is a rhombus.

3. AB is a straight line of given length: shew how to draw from A a line double the length of AB.

4. Two circles are drawn with the same centre 0, and two radii OA, OB are drawn in the smaller circle. If OA, OB are produced to cut the outer circle at D and E, prove that AD=BE.

5. AB is a straight line, and P, Q are two points, one on each side of AB. Shew how to find points in AB, whose distance from P is equal to PQ. How many such points will there be ?

6. In the figure of Proposition 2, if AB is equal to BC, shew that D, the vertex of the equilateral triangle, will fall on the circumference of the circle CGH.

7. In Proposition 2 the point A may be joined to either extremity of BC. Draw the figure, and prove the proposition in the case when A is joined to C.

8. On a given straight line AB describe an isosceles triangle having each of its equal sides equal to a given straight line PQ.

9. On a given base describe an isosceles triangle having each of its equal sides double of the base.

10. In a given straight line the points A, M, N, B are taken in order. On AB describe a triangle ABC, such that the side AC may be equal to AN, and the side BC to BM.

NOTE ON PROPOSITIONS 2 AND 3.

Propositions 2 and 3 are rendered necessary by the restriction tacitly imposed by Euclid, that compasses shall not be used to transfer distances. [See Notes on the Postulates.]

In carrying out the construction of Prop. 2 the point A may be joined to either extremity of the line BC; the equilateral triangle may be described on either side of the line so drawn ; and the sides of the equilateral triangle may be produced in either direction. Thus there are in general 2 x 2 x2, or eight, possible constructions. The student should exercise himself in drawing the various figures that

may arise.

PROPOSITION 4. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal, then the triangles shall be equal in all respects; that is to say, their bases or third sides shall be equal, and their remaining angles shall be equal, each to each, namely those to which the equal sides are opposite ; and the triangles shall be equal in area.

[blocks in formation]

Let ABC, DEF be two triangles, in which

the side AB is equal to the side DE,

the side AC is equal to the side DF, and the contained angle BAC is equal to the contained angle EDF. Then (i) the base BC shall be equal to the base EF;

(ii) the angle ABC shall be equal to the angle DEF;
(iii) the angle ACB shall be equal to the angle DFE;
(iv) the triangle ABC shall be equal to the triangle

DEF in area.
Proof. If the triangle ABC be applied to the triangle DEF,

so that the point A may lie on the point D,
and the straight line AB along the straight line DE;

then because AB is equal to DE, Нур. therefore the point B must coincide with the point E.

And because AB falls along DE, and the angle BAC is equal to the angle EDF, Hyp.

therefore AC must fall along DF.

And because AC is equal to DF, Hyp. therefore the point C must coincide with the point F.

Then since B coincides with E, and C with F, therefore the base BC must coincide with the base EF;

for if not, two straight lines would enclose a space; which is impossible.

Ax. 10. Thus the base BC coincides with the base EF, and is therefore equal to it.

Ax. 8. And the remaining angles of the triangle ABC coincide with the remaining angles of the triangle DEF, and are therefore equal to them; namely, the angle ABC is equal to the angle DEF,

and the angle ACB is equal to the angle DFE. And the triangle ABC coincides with the triangle DEF, and is therefore equal to it in area.

Ax. 8. That is, the triangles are equal in all respects. Q.E.D.

NOTE. The sides and angles of a triangle are known as its six parts. A triangle may also be considered in regard to its area.

Two triangles are said to be equal in all respects, or identically equal, when the sides and angles of one are respectively equal to the sides and angles of the other. We have seen that such triangles may be made to coincide with one another by superposition, so that they are also equal in area. [See Note on Axiom 8.]

[It will be shewn later that triangles can be equal in area without being equal in their several parts; that is to say, triangles can have the same area without having the same shape.]

EXERCISES ON PROPOSITION 4.

1. ABCD is a square : prove that the diagonals AC, BD are equal to one another.

2. ABCD is a square, and L, M, and N are the middle points of AB, BC, and CD: prove that (i) LM=MN.

(ii) AM=DM. (iii) AN=AM.

(iv) BN=DM. [Draw a separate figure in each case.] 3. ABC is an isosceles triangle : from the equal sides AB, AC two equal parts AX, AY are cut off, and BY and CX are joined. Prove that BY=CX.

4. ABCD is a quadrilateral having the opposite sides BC, AD equal, and also the angle BCD equal to the angle ADC : prove that BD is equal to AC.

[blocks in formation]
[blocks in formation]

The angles at the base of an isosceles triangle are equal to one another, and if the equal sides be produced, the angles on the other side of the base shall also be equal to one another.

[ocr errors][merged small][merged small]

I. 3.

Let ABC be an isosceles triangle, in which

the side AB is equal to the side AC, and let the straight lines AB, AC be produced to D and E. Then (i) the angle ABC shall be equal to the angle ACB;

(ii) the angle CBD shall be equal to the angle BCE. Construction. In BD take any point F; and from AE cut off a part AG equal to AF.

Join FC, GB.
Proof, Then in the triangles FAC, GAB,
FA is equal to GA,

Constr. and AC is equal to AB,

Нур. Because

also the contained angle at A is common to the

two triangles : therefore the triangle FAC is equal to the triangle GAB in all respects;

I. 4. that is, the base FC is equal to the base GB, and the angle ACF is equal to the angle ABG, also the angle AFC is equal to the angle AGB.

Again, because AF is equal to AG, and AB, a part of AF, is equal to AC, a part of AG; Hyp. therefore the remainder BF is equal to the remainder CG.

Then in the two triangles BFC, CGB,
BF is equal to CG,

Proved. and FC is equal to GB,

Proved. Because

also the contained angle BFC is equal to the contained angle CGB,

Proved. therefore the triangle BFC is equal to the triangle CGB in all respects;

I. 4. so that the angle FBC is equal to the angle GCB,

and the angle BCF to the angle CBG. Now it has been shewn that the angle ABG is equal to the

angle ACF, and that the angle CBG, a part of ABG, is equal to the angle

BCF, a part of ACF; therefore the remaining angle ABC is equal to the remaining angle ACB;

Ax. 3. and these are the angles at the base of the triangle ABC. Also it has been shewn that the angle FBC is equal to the

angle GCB; and these are the angles on the other side of the base. Q.E.D.

COROLLARY. Hence if a triongle is equilateral it is also equiangular.

[blocks in formation]
« ΠροηγούμενηΣυνέχεια »