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MISCELLANEOUS EXERCISËS ON SECTIONS İ. AND II.

71

MISCELLANEOUS EXERCISES ON SECTIONS I. AND II.

1. Shew that the construction in Proposition 2 may generally be performed in eight different ways. Point out the exceptional case.

2. The bisectors of two vertically opposite angles are in the same straight line.

3. In the figure of Proposition 16, if AF is joined, shew (i) that AF is equal to BC;

(ii) that the triangle ABC is equal to the triangle CFA in all respects.

4. ABC is a triangle right-angled at B, and BC is produced to D: shew that the angle ACV is obtuse.

5. Shew that in any regular polygon of n sides each angle contains 2(1–2)

right angles.

n

6. The angle contained by the bisectors of the angles at the base of any triangle is equal to the vertical angle together with half the sum of the base angles.

7. The angle contained by the bisectors of two exterior angles of any triangle is equal to half the sum of the two corresponding interior angles.

8. If perpendiculars are drawn to two intersecting straight lines from any point between them, shew that the bisector of the angle between the perpendiculars is parallel to (or coincident with) the bisector of the angle between the given straight lines.

9. If two points P, Q be taken in the equal sides of an isosceles triangle ABC, so that BP is equal to CQ, shew that PQ is parallel to BC.

10. ABC and DEF are two triangles, such that AB, BC are equal and parallel to DE, EF, each to each ; shew that AC is equal and parallel to DF.

11. Prove the second Corollary to Prop. 32 by drawing through any angular point lines parallel to all the sides.

12., If two sides of a quadrilateral are parallel, and the remaining two sides equal but not parallel, shew that the opposite angles are supplementary ; also that the diagonals are equal.

SECTION III.

THE AREAS OF PARALLELOGRAMS AND TRIANGLES.

Hitherto when two figures have been said to be equal, it has been implied that they are identically equal, that is, equal in all respects.

But figures may be equal in area without being equal in all respects, that is, without having the same shape.

The present section deals with parallelograms and triangles which are equal in area but not necessarily identically equal.

[The ultimate test of equality, as we have already seen, is afforded by Axiom 8, which asserts that magnitudes which may be made to coincide with one another are equal. Now figures which are not equal in all respects, cannot be made to coincide without first undergoing some change of form : hence the method of direct superposition is unsuited to the purposes of the present section.

We shall see however from Euclid's proof of Proposition 35, that two figures which are not identically equal, may nevertheless be so related to a third figure, that it is possible to infer the equality of their areas. ]

DEFINITIONS.

1. The Altitude of a parallelogram with reference to a given side as base, is the perpendicular distance between the base and the opposite side.

2. The Altitude of a triangle with reference to a given side as base, is the perpendicular distance of the opposite vertex from the base.

a

[From this point the following symbols will be introduced into the text:

= for is equal to ; .. for therefore. If it is thought desirable to shorten written work by the use of symbols and abbreviations, it is strongly recommended that only some well recognized system should be allowed, such, for example, as that given on page 11.]

PROPOSITION 35. THEOREM.
Parallelograms on the same base, and between the same
parallels, are equal in area.
DE
F A
DE

F A E DF

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B

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I. 34.

Let the parallelograms ABCD, EBCF be on the same base BC, and between the same parallels BC, AF.

Then shall the parallelogram ABCD be equal in area to the parallelogram EBCF.

CASE I. If the sides AD, EF, opposite to the base BC, are terminated at the same point D: then each of the parallelograms ABCD, EBCF is double of the triangle BDC;

.. they are equal to one another. Ax. 6. CASE II. But if the sides AD, EF are not terminated at the same point:

then because ABCD is a parallelogram,
... the side AD = the opposite side BC;
similarly EF = BC;
AD= EF.

Ax. 1.
:: the whole, or remainder, EA=the whole, or remainder, FD.
Then in the triangles FDC, EAB,
FD=EA,

Proved
and the side DC = the opposite side AB,
Because

also the exterior angle FDC = the interior opposite

angle EAB,

.. the triangle FDC = the triangle EAB. 1. 4. From the whole figure ABCF take the triangle FDC; and from the same figure take the equal triangle EAB;

then the remainders are equal. Ax. 3. Therefore the parallelogram ABCD is equal to the parallelo

I. 34.

I. 34.

I. 29.

gram EBCF.

PROPOSITION 36. THEOREM. Parallelograms on equal bases, and between the same parallels, are equal in area.

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Нур.

Let ABCD, EFGH be parallelograms on equal bases BC,
FG, and between the same parallels AH, BG.
Then shall the parallelogram ABCD be equal to the parallelo-

gram EFGH.
Construction. Join BE, CH.
Proof.

Then because BC = FG;
and the side FG=the opposite side EH;

I. 34. .:: BC=EH:

Ax. 1. and BC is parallel to EH;

Hyp. .:. BE and CH are also equal and parallel.

I. 33. Therefore EBCH is a parallelogram. Def. 36. Now the parallelograms ABCD, EBCH are on the same base BC, and between the same parallels BC, AH ; .. the parallelogram ABCD= the parallelogram EBCH. I. 35.

Also the parallelograms EFGH, EBCH are on the same base EH, and between the same parallels EH, BG; .. the parallelogram EFGH = the parallelogram EBCH. I. 35. Therefore the parallelogram ABCD is equal to the parallelogram EFGH.

Ax. 1. Q.E.D.

From the last two Propositions we infer that:

(i) A parallelogram is equal in area to a rectangle of equal base and equal altitude.

(ii) Parallelograms on equal bases and of equal altitudes are equal in area.

PROPOSITION 37. THEOREM. Triangles on the same base, and between the same parallels, are equal in area.

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Let the triangles ABC, DBC be upon the same base BC, and between the same parallels BC, AD.

Then shall the triangle ABC be equal to the triangle DBC.

Construction. Through B draw Be parallel to CA, to meet DA produced in E;

I. 31. through C draw CF parallel to BD, to meet AD produced in F.

Proof. Then, by construction, each of the figures EBCA, DBCF is a parallelogram.

Def. 36. And since they are on the same base BC, and between the

same parallels BC, EF; ;. the parallelogram EBCA= the parallelogram DBCF. 1. 35.

Now the diagonal AB bisects EBCA; I. 34. .. the triangle ABC is half the parallelogram EBCA. And the diagonal DC bisects DBCF;

I. 34. .. the triangle DBC is half the parallelogram DBCF.

And the halves of equal things are equal. Ax. 7. Therefore the triangle ABC is equal to the triangle DBC.

Q.E.D.

[For Exercises see page 79.]

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