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SECTION I.

On the Properties of Triangles.

PROPOSITION I. PROBLEM.

To describe an equilateral triangle on a given straight line.

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Let AB be the given st. line. It is required to describe an equilat. A on AB. With centre A and distance AB describe O BCD. Post. 3. With centre B and distance BA describe O ACE. Post. 3. From the pt. C, in which the Os cut one another, draw the st. lines CA, CB.

Post. 1.
Then will ABC be an equilat. A.
For

::: A is the centre of O BCD,
.:. AC=AB.

Def. 13.
And :: B is the centre of O ACE,
BC=AB.

Def. 13. Now :: AC, BC are each=AB, .. AC=BC.

Ax. 1. Thus AC, AB, BC are all equal, and an equilat. A ABC has been described on AB.

Q. E. F.

PROPOSITION II. PROBLEM.

From a given point to draw a straight line equal to a given straight line.

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B

Let A be the given pt., and BC the given st. line.

It is required to draw from A a st. line equal to BC.

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From A to B draw the st. line AB.

Post. 1. On AB describe the equilat. A ABD.

I. 1. With centre B and distance BC describe O CG H. Post. 3.

Produce DB to meet the Oce CGH in G. With centre D and distance DG describe O GKL. Post. 3.

Produce DA to meet the Oce GKL in L.

Then will AL=BC.
For

::: B is the centre of O CGH,
.:. BC=BG.

Def. 13. And :: D is the centre of O GKL, .: DL=DG.

Def. 13. And parts of these, DA and DB, are equal. Def. 21.

... remainder AL=remainder BG. Ax. 3. But BC=BG; .. AL=BC.

Ax. l. Thus from pt. A a st. line AL has been drawn=BC.

Q. E, F.

PROPOSITION III. PROBLEM. From the greater of two given straight lines to cut off a part equal to the less.

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Let AB be the greater of the two given st. lines AB, CD.

It is required to cut off from AB a part=CD.
From A draw the st. line AE=CD.

I. 2. With centre A and distance A.E describe o EFH,

cutting AB in F. Then will AF=CD. For

:: A is the centre of O EFH,

.. AF=AE. But

AE=CD; .. AF=CD.

Ax. 1. Thus from AB a part AF has been cut off=CD.

Q. E. F. EXERCISES. 1. Shew that if raight lines be drawn from A and B in the diagram of Prop. I. to the other point in which the circles intersect, another equilateral triangle will be described on AB.

2. By a construction similar to that in Prop. 111. produce the less of two given straight lines that it may be equal to the greater.

3. Draw a figure for the case in Prop. II., in which the given point coincides with B.

4. By a similar construction to that in Prop. I. describe on a given straight line an isosceles triangle, whose equal sides shall be each equal to another given straight line.

PROPOSITION IV. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, they must have their third sides equal ; and the two triangles must be equal, and the other angles must be equal, each to each, viz. Those to which the equal sides are opposite.

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In the As ABC, DEF, let AB=DE, and AC=DF, and 2 BAC= EDF. Then must BC=EF and A ABC = A DEF, and the other Ls, to which the equal sides are opposite, must be equal, that is, L ABC= _ DEF and LACB= 2 DFE.

For, if ABC be applied to A DEF, so that A coincides with D, and AB falls on DE, then : AB=DE, .. B will coincide with E. And :: AB coincides with DE, and BAC= . EDF, Hyp

.. AC will fall on DF.
Then ::: AC=DF, .-. C will coincide with F.
And :: B will coincide with E, and C with F,

.: BC will coincide with EF; for if not, let it fall otherwise as EOF: then the two st. lines BC, EF will enclose a space, which is impossible. Post. 5.

:: BC will coincide with and .. is equal to EF, Ax. 8. and A ABC..

A DEF, and L ABC..

- DEF, and 4 ACB..

DFE.

Q. E. D.

NOTE 1. On the Method of Superposition. Two geometrical magnitudes are said, in accordance with Ax. viii. to be equal, when they can be so placed that the boundaries of the one coincide with the boundaries of the other.

Thus, two straight lines are equal, if they can be so placed that the points at their extremities coincide : and two angles are equal, if they can be so placed that their vertices coincide in position and their arms in direction : and two triangles are equal, if they can be so placed that their sides coincide in direction and magnitude.

In the application of the test of equality by this Method of Superposition, .we assume that an angle or a triangle may be moved from one place, turned over, and put down in another place, without altering the relative positions of its boundaries.

We also assume that if one part of a straight line coincide with one part of another straight line, the other parts of the lines also coincide in direction ; or, that straight lines, which coincide in two points, coincide when produced.

The method of Superposition enables us also to compare magnitudes of the same kind that are unequal. For example, suppose ABC and DEF to be two given angles.

B

F Suppose the arm BC to be placed on the arm EF, and the vertex B on the vertex E.

Then, if the arm BA coincide in direction with the arm ED, the angle ABC is equal to DEF.

If BA fall between ED and EF in the direction EP, ABC is less than DEF.

If BA fall in the direction EQ so that ED is between EQ and EF, ABC is greater than DEF.

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