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11. Prove the second part of the proposition by drawing through A a straight line DAE || BC. (The Pythagorean proof.)

12. If any of the angles of an isosceles triangle be two-thirds of a right angle, the triangle must be equilateral.

13. Each of the base angles of an isosceles triangle equals half the exterior vertical angle.

14. If the exterior vertical angle of an isosceles triangle be bisected, the bisector is || the base.

15. Show that the space round a point can be filled up with six equilateral triangles, or four squares, or three regular hexagons. 16. Can a right angle be divided into any other number of equal parts than two or three?

17. In a right-angled triangle, if a perpendicular be drawn from the

right angle to the hypotenuse, the triangles on each side of it are equiangular to the whole triangle and to one another. 18. Prove the seventh deduction indirectly; and also directly by producing the median to the hypotenuse its own length. 19. If the arms of one angle be respectively perpendicular to the arms of another, the angles are either equal or supplementary. 20. Prove Cor. 3 by taking a point inside the figure and joining it to the angular points.

PROPOSITION 33. THEOREM.

The straight lines which join the ends of two equal and *parallel straight lines towards the same parts, are themselves equal and parallel.

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Let AB and CD be equal and parallel:

it is required to prove AC and BD equal and parallel.

Join BC.

Because BC meets the parallels AB, CD,

.. L ABC = alternate

DCB.

I. 29

[blocks in formation]

Proved

Because CB meets AC and BD, and makes the alter

nate LS ACB, DBC equal;

.. AC is || BD.

1. State a converse of this proposition.

I. 27

2. If a quadrilateral have one pair of opposite sides equal and parallel, it is a ||m.

3. What statements may be made about the straight lines which join the ends of two equal and parallel straight lines towards opposite parts?

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A parallelogram has its opposite sides and angles equal, and is bisected by either diagonal.

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Let ACDB be a ||m of which BC is a diagonal :

it is required to prove that the opposite sides and angles of ACDB are equal, and that ▲ ABC = ▲ DCB.

Because BC meets the parallels AB, CD,

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and because BC meets the parallels AC, BD,

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I. 29

I. 29 Proved

=

[blocks in formation]

L ABC = L DCB

LACB

BC = CB;

A

C

... AB = DC, AC = DB, L BAC LCDB,

D

=

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COR. If the arms of one angle be respectively parallel to the arms of another, the angles are either (1) equal or (2) supplementary.

For (1) BAC has been proved = L CDB ;

and (2) if BA be produced to E,

LEAC, which is supplementary to BAC,

must be supplementary to CDB.

I. 13

1. If two sides of a m which are not opposite to each other be equal, all the sides are equal.

2. If two angles of a ||m which are not opposite to each other be equal, all the angles are right.

3. If one angle of a ||m be right, all the angles are right.

4. If two ||ms have one angle of the one = one angle of the other, the ||ms are mutually equiangular.

5. If a quadrilateral have its opposite sides equal, it is a ||m.

6. If a quadrilateral have its opposite angles equal, it is a ||m.

7. If the diagonals of a m be equal to each other, the ||m is a rectangle.

8. If the diagonals of a m bisect the angles through which they pass, the m is a rhombus.

9. If the diagonals of a ||m cut each other perpendicularly, the ||m is a rhombus.

10. If the diagonals of a m be equal and cut each other perpendicularly, the ||m is a square.

11. Show how to bisect a straight line by means of a pair of parallel

rulers.

12. Every straight line drawn through the intersection of the diagonals of a ||m, and terminated by a pair of opposite sides, is bisected, and bisects the m.

13. Bisect a given ||m by a straight line drawn through a given point either within or without the [m.

=

14. The straight line joining the middle points of any two sides of a triangle is || the third side and half of it. 15. If the middle points of the three sides of a triangle be joined with each other, the four triangles thence resulting are equal. 16. Construct a triangle, having given the middle points of its three sides.

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Parallelograms on the same base and between the same parallels are equal in area.

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Let ABCD, EBCF bem on the same base BC, and between the same parallels AF, BC:

it is required to prove || ABCD

=

Because AF meets the parallels AB, DC,

exterior FDC;

.. interior A =
and because AF meets the parallels EB, FC,

[blocks in formation]

|| EBCF.

I. 29

[blocks in formation]

In As ABE, DCF, L AEB

=

L DFC

I. 29 Proved Proved

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מנון

=

NOTE. This proposition affords a means of measuring the area of a m; thence (by I. 34 or 41) the area of a triangle; and thence (by I. 37, Cor.) the area of any rectilineal figure. For the area of any the area of a rectangle on the same base and between the same parallels ; and it is, or ought to be, explained in books on Mensuration, that the area of a rectangle is found by taking the product of its length and breadth. This phrase 'taking the product of its length and breadth,' means that the numbers, whether integral or not, which express the length and breadth in terms of the same linear unit, are to be multiplied together. Hence the method of finding the area of a ||m is to take the product of its base and altitude, the altitude being defined to be the perpendicular drawn to its base from any point in the side opposite.

1. Prove the proposition for the case when the points D and E coincide.

2. Equal ms on the same base and on the same side of it are between the same parallels.

3. If through the vertices of a triangle straight lines be drawn || the opposite sides, and produced till they meet, the resulting figure will contain three equal ||ms.

4. On the same base and between the same parallels as a given ||m, construct a rhombus = them.

5. Prove the equality of ▲s ABE and DCF in the proposition by I. 4 (as Euclid does), or by I. 8, instead of by I. 26.

PROPOSITION 36. THEOREM.

Parallelograms on equal bases and between the same parallels are equal in area.

[blocks in formation]

Let ABCD, EFGH be ||ms on equal bases BC, FG, and

between the same parallels, AH, BG:

it is required to prove || ABCD = || EFGH.

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