Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

C

B

K

Let AB be a straight line divided into any two parts in C; upon AB describe the square ADEB (42); join BD; through the point C draw CGF parallel to AD or BE, and meeting BDin G; and through H G draw HGK parallel to AB. Then, since BCGK is a parallelogram by construction, its opposite sides are equal to each other, and likewise its opposite angles (40), that is, BC= KG, BK = CG, 4 CBK: = < CGK, and BCG: =< BKG. But since BE = ED, .. ‹ EBD=¿ EDB (26); and since KG is parallel to ED, ≤ KGB = ▲ EDB (34), .. ≤ KBG, which is the same as ‹ EBD, = < KGB, and ... BK = KG; but BC= KG, and BK = CG, .. BCGK is equilateral.

==

[ocr errors]

D

F

[ocr errors]

Again, since BC is parallel to KG, CBK+ < BKG = two right angles (34); but ▲ CBK is a right angle, .. BKG is also a right angle; and the opposite angles are equal, .. BCGK has all its angles right angles. And it has been proved to have all its sides equal. It is therefore a square; and it is the square of BC.

Similarly it may be shewn, that HGFD is a square; and it is the square of HG, or AC, since ACGH is a parallelogram of which the side AC- HG.

=

Also, since BCG is a right angle, .. ACG is a right angle, and the parallelogram ACGH is a rectangle, and it is 'contained by* AC, CG, or by AC, CB, since CB= CG. And, similarly, EFGK is a rectangle, contained by FG, GK, or by HG, GC, or by AC, CB. And these make up the whole area ABED, which is the square of AB; .. the square of AB= the square of AC + the square BC + twice the rectangle AC, CB †.

(40), and since each of the angles of a rectangle is always a right angle, two adjacent sides alone will obviously serve to fix any rectangle; and hence it is, that the rectangle is said to be 'contained' by those sides, because nothing more is needed to determine the rectangle.

The expression 'contained by' is mostly, omitted; and the rectangle contained by any two lines, as AC, and CB, is simply called 'the rectangle AC, CB'.

By the help of this Proposition a very elegant proof of the important Theorem ́in (43) may be given as follows:

45. PROP. XXIII. If a straight line be divided into any two parts, the squares of the whole line and one of the parts are together equal to twice the rectangle contained by the whole and that part, together with the square of the other part.

=

=

With the same construction as in (44), and in the same manner it may be shewn, that BCGK, and HGFD, are the squares of BC, and AC; and that ACGH is a rectangle, and EFGK, which is also a rectangle. Add to each of these equals the square BCGK, and then the rectangle ABKH = the rectangle EBCF; .. ABKH+ EBCF twice the rectangle ABKH = twice the rectangle contained by AB, BC. Now add to these equals the square HDFG, which is the square of AC; then ABKH + EBCF+ square of AC=twice the rectangle AB, BC + square of AC. But the former of these equals make up the square ABED + the square BCGK; the squares of AB and BC are together equal to twice the rectangle AB, BC, together with the square of AC.

46. PROP. XXIV. In any obtuse-angled triangle if a perpendicular be drawn from the vertex of either of the acute angles upon the opposite side produced, the square of the side subtending the obtuse angle is greater than the sum of the squares of the sides forming the obtuse angle by twice the rectangle contained by the side which is produced and the part produced, viz. the part intercepted between the perpendicular and the vertex of the obtuse angle.

Let ACB be an obtuse angle of the triangle ABC; and from A draw AD perpendicular to BC produced;

Let ABC be a right-angled triangle; C the right angle; produce CA to D, making AD=CB. Upon CD describe the square DCEF. Take EG= CB, and FH

CB; and join BG, GH, HA. It may then easily be shewn, that ABGH is a square, and that the four triangles ABC, BGE, GHF, HAD are equal to one another in all respects, so that the sum of them is equal to twice the rectangle AC, CB, or AC, AD, since AD = CB. Hence the square of CD=the square of AB + twice

H

F

G E

the rectangle AC, AD. But by (44) the square of CD=the square of AC+ square of AD+twice the rectangle AC, AD. Therefore the square of AB=square of AC+ square of AD= square of AC + square of CB.

[ocr errors][merged small]

the square of AB shall be greater than the sum of the squares of AC, and BC, by twice

the rectangle BC, CD.

For the square of BD = the square of BC + the square of CD + twice the rectangle BC, CD (44). Add to these equals the square of AD; then the square of BD+ square of AD

=

D

square of BC+ square of CD + square of AD + twice the rectangle BC, CD. But (43) square BD+ square of AD= square of AB; and square of CD + square of AD = square of AC; • square of AB square of BC+ square of AC+twice the rectangle BC, CD; that is, square of AB is greater than the sum of the squares of AC, and BC, by twice the rectangle BC, CD.

=

47. PROP. XXV. In every triangle the square of the side opposite to any acute angle is less than the sum of the squares of the sides forming that angle by twice the rectangle contained by either of these sides and that part of it which is intercepted between a perpendicular let fall upon it from the vertex of the opposite angle and the acute angle.

Let ABC be an acute angle of the triangle ABC; from A draw AD perpendicular to BC meeting it in the point D. Then the square of AC shall be less than the squares of AB and BC by twice the rectangle BC, BD.

=

For by (45) the square of BC+the square of BD = twice the rectangle BC, BD+the square of CD. Add to each of these equals the square of AD; then the square of BC + the square of BD + the square of

AD twice the rectangle BC, B D

=

BD + the square of CD+ the square of AD. But, since ADB is a right angle, the squares of BD, and AD, are equal to the square of AB (43); also the squares of CD and AD are equal to the square of AC; . the square of BC + the square of AB= twice the rectangle BC, BD+ the square of AC; that is, the square of AC is less than the sum of the squares of AB, and BC by twice the rectangle BC, BD.

QUESTIONS AND EXERCISES IN THE PRECEDING

PROPOSITIONS. A.

(1) In describing an equilateral triangle (23) upon a given straight line, how much is taken for granted? If a second triangle be drawn on the opposite side of the line by a similar construction, what figure will the two together make?

(2) Have we yet laid down any mode of measuring an angle? If not, how are we able to prove that one angle is equal to, less than, or greater than, another according to circumstances?

(3) If the angles of one triangle be equal to the angles of another, each to each, are the triangles necessarily equal? What is the force of the expression, 'each to each'? Exhibit a case where the angles are respectively equal, but not 'each to each'.

(4) What is meant in (25) by a 'given angle'? Is it necessary that the triangle DEF should be equilateral? What other triangle would do as well? Does it matter on which side of DE the triangle is described?

(5) Define an 'isosceles' triangle. Is an equilateral triangle isosceles? Can more than one isosceles triangle be constructed on the same base and on the same side of it? Can a right-angled triangle be also isosceles ?

(6) What is the precise meaning of 'given straight line' in (27), where it is required to bisect it? Is it the same as in (28) and (29)? If not, what is the dif

ference?

(7) What is the meaning of the word 'base' as applied to a triangle, and to a parallelogram? Is it restricted to one fixed side only?

(8) Shew that only one straight line can be drawn perpendicular to a given straight line from a given point without it.

(9) Shew that the perpendicular is the shortest of all lines from a given point to a given straight line. Of all such lines which measures the distance of the point from the given line?

(10) If in (30) straight lines be drawn bisecting

each of the angles ACD, BCD, shew that these straight lines are at right angles to one another.

(11) Shew that any point in the straight line bisecting an angle is equidistant from the two straight lines forming the angle.

(12) Shew that any side of a triangle is less than half the sum of all three sides of the same triangle.

(13) Shew that the straight line drawn from the middle point of the base of an isosceles triangle to the vertex of the opposite angle is at right angles to the base, and bisects the opposite angle.

(14) Shew that each angle of an equilateral triangle is two-thirds of a right angle. Trisect a right angle.

(15) Can a triangle have more than one of its angles a right angle, or an obtuse angle? If not, why not?

(16) Shew that the four angles of every quadrilateral figure are together equal to four right angles.

(17) If one of the angles of a parallelogram be a right angle, does this determine all the other angles?

(18) Shew that any two straight lines at right angles to the same straight line, and on the same side of it, are parallel.

(19) If two parallel straight lines be intersected by two other parallel straight lines, shew that the parts of the latter two intercepted between the former two are equal to each other.

(20) If two straight lines in the same plane be equal and parallel, shew that the straight lines joining their extremities towards the same parts are also equal and parallel.

(21) If two straight lines be drawn bisecting two angles of a triangle, shew that the point in which they intersect is equidistant from the three sides of the triangle.

(22) Is it correct to speak of drawing a line from an angle? The expression is found in Simson's Euclid; what does it mean? See definition of angle.

(23) Simson also, after defining 'vertex' of an angle, on the first occasion of using the term (Prop. vII) speaks

« ΠροηγούμενηΣυνέχεια »