EXERCISES AND RIDERS ON PROPOSITIONS 33 AND 34.

1. Find the locus of a point which is always at the same distance from a given straight line.

2. What would be the difference in the enunciation of Prop. 33 if the extremities of the equal and parallel straight lines were joined, but not towards the same parts ?

3. Name and define the four different kinds of parallelograms.

4. The diagonals of a parallelogram bisect each other.

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

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

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

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

9. When are the diagonals of a parallelogram equal?

10. When does the diagonal of a parallelogram bisect the opposite angles? 11. If a quadrilateral figure is equilateral and has one angle a right angle, then all its angles are right angles.

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

13. Prove that any straight line drawn through the middle point M of a diagonal of a parallelogram and terminated by opposite sides of the parallelogram is bisected at M.

14. If a quadrilateral figure has two of its opposite sides parallel and the other two sides equal but not parallel, then any two opposite angles of the quadrilateral are together equal to two right angles.

15. ABCD is a parallelogram and BD is a diagonal. Prove that the perpendiculars from A and C on BD are equal.

16. Bisect a parallelogram by a straight line drawn (1) through a given point in one of its sides; or (2) parallel to a given straight line; or (3) perpendicular to two of its opposite sides.

17. In the figure of Prop. 34, if CA is produced to E so that AE is equal to CA, and if the parallelogram BAEF is completed, show that DA and AF are in one and the same straight line.

18. Construct a triangle whose angles shall be equal to those of a given triangle and whose area shall be four times the area of the given triangle.

PART III.

THE AREAS OF PARALLELOGRAMS, TRIANGLES AND SQUARES.

The word "equal" is used in two senses in Euclid. In the first part of Euclid it is generally used to mean equal in all respects. The test of such equality is as stated in Axiom 8, that the given magnitudes can be made to coincide, that is, to exactly fit on one another. Such equality is specially applicable to lines and angles; but it is possible for figures to be equal in area only, without having the lines by which they are bounded equal in length. Throughout the rest of Book I. and in Book II., Euclid therefore often uses the word “equal" to mean “equal in area.” The test of this second kind of equality is that all the parts of the area of one figure may be shifted in such a way that they would exactly fit the area of another figure. When this is possible the figures are equal in area.

DEFINITIONS.

1. The altitude or height of a parallelogram is the perpendicular distance between two of its opposite sides.

2. The altitude or height of a triangle is the perpendicular distance of any one of its angular points from the opposite side.

PROP. 35.-Theorem.-Parallelograms on the same base and between the same parallels are equal in area.

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

the parallelogram ABCD shall be equal to the parallelogram

EBCF in area.

Proof.-CASE I.-When the point D coincides with E. Each of the parallelograms ABCD, EBCF is double of the triangle

therefore the parallelogram ABCD is equal in area to the

parallelogram EBCF.

CASE II. When the point D does not coincide with E.
(1) Because ABCD is a parallelogram,

therefore AD is equal to BC.

And because EBCF is a parallelogram,

therefore EF is equal to BC.

Therefore AD is equal to EF.

Prop. 34

Prop. 34

Ax. 1

Add or subtract DE;

therefore the whole or remainder AE is equal to the whole or

Because

remainder DF.

(2) Again, because in the two triangles FDC, EAB,

and the exterior angle FDC is equal to its interior

opposite angle EAB;

Prop. 29 therefore by Proposition 4 the triangle FDC is equal to the

triangle EAB in area.

(3) From the whole figure ABCF take the triangle FDC, and from the whole figure ABCF take the equal triangle EAB, then, the remainders are equal in area;

that is, the parallelogram ABCD is equal in area to the

parallelogram EBCF.

Ax. 3

Q. E. D.

PROP. 36.-Theorem.-Parallelograms on equal bases and between the same parallels are equal in area.

Let the parallelograms ABCD, EFGH be on equal bases BC, FG, and between the same parallels AH, BG;

the parallelogram ABCD shall be equal to the parallelogram EFGH in area.

Again, because the parallelograms ABCD and EBCH are on the same base BC and between the same parallels BC and AH, therefore the parallelogram ABCD is equal to the parallelogram EBCH in area.

Prop. 35 And because the parallelograms EFGH and EBCH are on the same base EH and between the same parallels EH and BG; therefore the parallelogram EFGH is equal to the parallelogram

EBCH in area.

Prop. 35

Therefore the parallelogram ABCD is equal to the parallelogram

EFGH in area.

Ax. 1

EXERCISE.

Q. E. D.

If a rectangle ABCD and a rhombus EFGH be on equal bases and have equal areas, show that the perimeter of the rectangle is less than the perimeter of the rhombus.

PROP. 37.-Theorem.-Triangles on the same base and between the same parallels are equal in area.

Let the triangles ABC, DBC be on the same base BC and
between the same parallels AD and BC;

the triangle ABC shall be equal to the triangle DBC in area.

Construction. Produce AD both ways to E and F.
Through B draw BE parallel to AC,

and through C draw CF parallel to BD.

Prop. 31

Proof. Because EBCA and DBCF are parallelograms on the same base BC and between the same parallels BC and EF, therefore the parallelogram EBCA is equal to the parallelogram

DBCF in area.

Prop. 35 But the triangle ABC is half of the parallelogram EBCA, Prop. 34 and the triangle DBC is half of the parallelogram DBCF, Prop. 34 and the halves of equal things are equal;

Ax. 7 therefore the triangle ABC is equal to the triangle DBC in area.

Q. E. D.

EXERCISES.

1. In the figure of Prop. 37 show that the triangles ABD and ACD are equal in area.

2. ABC is a given triangle; show how to construct a right-angled triangle equal in area to ABC.

3. ABC is a given triangle, and DE a given straight line which is not parallel to BC; show how to construct on AB a triangle equal in area to ABC and having its vertex in DE.