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PROPOSITION 34. THEOREM.

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelo

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Let ACDB be a parallelogram, of which BC is a diagonal: then shall the opposite sides and angles of the figure be equal to one another; and the diagonal BC shall bisect it.

Proof.

them,

Because AB and CD are parallel, and BC meets

therefore the alternate angles ABC, DCB are equal. 1. 29. Again, because AC and BD are parallel, and BC meets them,

therefore the alternate angles ACB, DBC are equal. 1. 29. Hence in the triangles ABC, DCB,

Because

(the angle ABC is equal to the angle DCB,
and the angle ACB is equal to the angle DBC;
also the side BC, which is adjacent to the equal
angles, is common to both,

therefore the two triangles ABC, DCB are equal in all respects;

I. 26.

so that AB is equal to DC, and AC to DB; and the angle BAC is equal to the angle CDB. Also, because the angle ABC is equal to the angle DCB, and the angle CBD equal to the angle BCA, therefore the whole angle ABD is equal to the whole angle DCA.

And since it has been shewn that the triangles ABC, DCB are equal in all respects,

therefore the diagonal BC bisects the parallelogram ACDB.

[See note on next page.]

Q. E. D.

NOTE. To the proof which is here given Euclid added an application of Proposition 4, with a view to shewing that the triangles ABC, DCB are equal in area, and that therefore the diagonal BC bisects the parallelogram. This equality of area is however sufficiently established by the step which depends upon 1. 26. [See page 48.]

EXERCISES.

1. If one angle of a parallelogram is a right angle, all its angles are right angles.

2. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram.

3. If the opposite angles of a quadrilateral are equal, the figure is a parallelogram.

4. If a quadrilateral has all its sides equal and one angle a right angle, all its angles are right angles; that is, all the angles of a square are right angles.

5. The diagonals of a parallelogram bisect each other.

6. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram.

7. If two opposite angles of a parallelogram are bisected by the diagonal which joins them, the figure is equilateral.

8. If the diagonals of a parallelogram are equal, all its angles are right angles.

9. In a parallelogram which is not rectangular the diagonals are unequal.

10. Any straight line drawn through the middle point of a diagonal of a parallelogram and terminated by a pair of opposite sides, is bisected at that point.

11. If two parallelograms have two adjacent sides of one equal to two adjacent sides of the other, each to each, and one angle of one equal to one angle of the other, the parallelograms are equal in all respects.

12. Two rectangles are equal if two adjacent sides of one are equal to two adjacent sides of the other, each to each.

13. In a parallelogram the perpendiculars drawn from one pair of opposite angles to the diagonal which joins the other pair are equal.

14. If ABCD is a parallelogram, and X, Y respectively the middle points of the sides AD, BC; shew that the figure AYCX is a parallelogram.

MISCELLANEOUS EXERCISES ON SECTIONS I. AND II.

65

MISCELLANEOUS EXERCISES ON SECTIONS I. AND II.

1. Shew that the construction in Proposition 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 ACD is obtuse.

5. Shew that in any regular polygon of n sides each angle contains 2(n-2) right angles.

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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.

H. E.

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.

In Section III. of Euclid's first Book, we have to consider the equality in area of parallelograms and triangles which are not necessarily equal in all respects.

[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 identically equal, 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.

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Parallelograms on the same base, and between the same

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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 of the given parallelograms, opposite to the common base BC, are terminated at the same point D:

then because each of the parallelograms is double of the triangle BDC;

therefore they are equal to one another.

I. 34.

Ax. 6.

CASE II. But if the sides AD, EF, opposite to the base BC, are not terminated at the same point:

then because ABCD is a parallelogram, therefore AD is equal to the opposite side BC; and for a similar reason, EF is equal to BC; therefore AD is equal to EF.

I. 34.

Ax. 1.

Hence the whole, or remainder, EA is equal to the whole,

or remainder, FD.

Because

Then in the triangles FDC, EAB,

FD is equal to EA,

Proved.

and DC is equal to the opposite side AB, 1. 34. also the exterior angle FDC is equal to the interior opposite angle EAB,

I. 29.

therefore the triangle FDC is equal to 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. that is, the parallelogram ABCD is equal to the parallelogram EBCF.

Q. E. D.

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