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EXERCISES ON PROPOSITIONS 12-26.

1. If BX and CY, the bisectors of the angles at the base BC of an isosceles triangle ABC, meet the opposite sides in X and Y, shew that the triangles YBC, XCB are equal in all respects.

2. Shew that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.

3. Any point on the bisector of an angle is equidistant from the arms of the angle.

4. Through O, the middle point of a straight line AB, any straight line is drawn, and perpendiculars AX and BY are dropped upon it from A and B : shew that AX is equal to BY.

5. If the bisector of the vertical angle of a triangle is at right angles to the base, the triangle is isosceles.

6. The perpendicular is the hortest straight line that can be drawn from a given point to a given straight line ; and of others, that which is nearer to the perpendicular is less than the more remote ; and two, and only two equal straight lines can be drawn from the given point to the given straight line, one on each side of the perpendicular.

7. From two given points on the same side of a given straight line, draw two straight lines, which shall meet in the given straight line, and make equal angles with it.

Let AB be the given straight line, and P, Q the given points.

It is required to draw from P and Q to a point in AB, two straight lines

A
к

ІН В that shall be equally inclined to AB. Construction. From P draw PH

P perpendicular to AB: produce PH to P, making HP equal to PH. Draw QP', meeting AB in K. Join PK.

Then PK, QK shall be the required lines. [Supply the proof.]

8. In a given straight line find a point which is equidistant from two given intersecting straight lines. In what case is this impossible?

9. Through a given point draw a straight line such that the perpendiculars drawn to it from two given points may be equal.

In what case is this impossible ?

SECTION II.

PARALLEL STRAIGHT LINES AND PARALLELOGRAMS.

1/2 4/3

DEFINITION. Parallel straight lines are such as, being in the same plane, do not meet however far they are produced in both directions.

When two straight lines AB, CD are met by a third straight line EF, eight angles are formed, to which for the sake of distinction particular names

E

A are given. Thus in the adjoining figure,

B 1, 2, 7, 8 are called exterior angles, 3, 4, 5, 6 are called interior angles,

5/6 4 and 6 are said to be alternate angles ;

8/7 so also the angles 3 and 5 are alternate to one another.

Of the angles 2 and 6, 2 is referred to as the exterior angle, and 6 as the interior opposite angle on the same side of EF.

2 and 6 are sometimes called corresponding angles.
So also, 1 and 5, 7 and 3, 8 and 4 are corresponding angles.

Euclid's treatment of parallel straight lines is based upon his twelfth Axiom, which we here repeat.

AXIOM 12. If a straight line cut two straight lines so as to make the two interior angles on the same side of it together less than two right angles, these straight lines, being continually produced, will at length meet on that side on which are the angles which are together less than two right angles.

Thus in the figure given above, if the two angles 3 and 6 are together less than two right angles, it is asserted that AB and CD will meet towards B and Ď.

This Axiom is used to establish 1. 29 : some remarks upon it will be found in a note on that Proposition.

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If a straight line, falling on two other str eight lines, make the alternate angles equal to one another, then these two straight lines shall be parallel.

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Let the straight line EF cut the two straight lines AB, CD at G and H, so as to make the alternate angles AGH, GHD equal to one another.

Then shall AB and CD be parallel. Proof. For if AB and CD be not parallel, they will meet, if produced, either towards B and D, or

towards A and c. If possible, let AB and CD, when produced, meet towards B

and D, at the point K. Then KGH is a triangle, of which one side KG is produced

to A; therefore the exterior angle AGH is greater than the interior opposite angle GHK.

I. 16. But the angle AGH was given equal to the angle GHK: Hyp. hence the angles AGH and GHK are both equal and unequal;

which is impossible. Therefore AB and CD cannot meet when produced towards

B and D. Similarly it may be shewn that they cannot meet towards A and C:

therefore AB and CD are parallel.

PROPOSITION 28.

THEOREM. If a straight line, falling on two other straight lines, make an exterior angle equal to the interior opposite angle on the same side of the line ; or if it make the interior angles on the same side together equal to two right angles, then the two straight lines shall be parallel.

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I. 27.

Let the straight line EF cut the two straight lines AB, CD in G and H: and

First, let the exterior angle EGB be equal to the interior opposite angle GHD.

Then shall AB and CD be parallel. Proof. Because the angle EGB is equal to the angle GHD; and because the angle EGB is also equal to the vertically opposite angle AGH ;

1. 15. therefore the angle AGH is equal to the angle GHD;

but these are alternate angles ; therefore AB and CD are parallel.

Q.E.D. Secondly, let the two interior angles BGH, GHD be together equal to two right angles.

Then shall AB and CD be parallel. Proof. Because the angles BGH, GHD are together equal to two right angles ;

Hyp. and because the adjacent angles BGH, AGH are also together equal to two right angles;

I. 13. therefore the angles BGH, AGH are together equal to the two angles BGH, GHD.

From these equals take the common angle BGH : then the remaining angle AGH is equal to the remaining angle GHD: and these are alternate angles ; therefore AB and CD are parallel.

I. 27. Q.E.D.

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PROPOSITION 29. THEOREM. If a straight line fall on two parallel straight lines, then shall make the alternate angles equal to one another, and the exterior angle equal to the interior opposite angle on the same side ; and also the two interior angles on the same side equal to two right angles.

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Let the straight line EF fall on the parallel straight lines AB, CD. Then (i) the angle AGH shall be equal to the alternate angle

GHD; (ii) the exterior angle EGB shall be equal to the interior

opposite angle GHD; (iii) the two interior angles BGH, GHD shall be together

equal to two right angles. Proof. (i) For if the angle AGH be not equal to the angle

GHD, one of them must be greater than the other. If possible, let the angle AGH be greater than the angle GHD;

add to each the angle BGH : then the angles AGH, BGH are together greater than the

angles BGH, GHD. But the adjacent angles AGH, BGH are together equal to

two right angles ; therefore the angles BGH, GHD are together less than two

right angles; therefore, by Axiom 12, AB ard CD meet towards B and D.

But they never meet, since they are parallel. Hyp. Therefore the angle AGH is not unequal to the angle GHD: that is, the angle AGH is equal to the alternate angle GHD.

(Over)

I. 13.

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