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PROPOSITION 11. PROBLEM. To divide a given straight line into two parts, so that the rectangle contained by the whole and one part may be equal to the square on the other part.

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I. 3.

Let AB be the given straight line. It is required to divide AB into two parts, so that the rectangle contained by the whole and one part may be equal to the square on the other part. Construction. On AB describe the square ACDB.

I. 46.
Bisect AC at E.

I. 10.
Join EB.
Produce CA to F, making EF equal to EB.
On AF describe the square AFGH.

I. 46. Then shall AB be divided at H, so that the rect. AB, BH is equal to the sq. on AH.

Produce GH to meet CD in K. Proof. Because CA is bisected at E, and produced to F, .. the rect. CF, FA with the sq. on EA= the sq. on EF

II. 6. = the

sq. on EB. Constr. But the sq. on EB= the sum of the sqq. on EA, AB, for the angle EAB is a rt. angle.

I. 47. .. the rect. CF, FA with the sq. on EA= the sum of the sqq. on EA, AB.

From these equals take the sq. on EA:
then the rect. CF, FA= the sq. on AB.

But the rect. CF, FA= the fig. FK; for FA= FG;
and the sq. on AB= the fig. AD.

Constr.
.. the fig. FK = the fig. AD.
From these equals take the common fig. AK;
then the remaining fig. FH=the remaining fig. HD.
But the fig. HD=the rect. AB, BH; for BD = AB;

and the fig. FH is the sq. on AH.

... the rect. AB, BH=the sq. on AH. Q.E.F. DEFINITION. A straight line is said to be divided in Medial Section when the rectangle contained by the given line and one of its segments is equal to the square on the other segment.

The student should observe that this division may be internal or external.

Thus if the straight line AB is divided internally at H, and ex. ternally at H', so that (i) AB. BH =AH, H'

A

H B (ii) AB. BH'=AH', we shall in either case consider that AB is divided in medial section.

The case of internal section is alone given in Euclid 11. 11; but a straight line may be divided externally in medial section by a similar process. See Ex. 21, p. 160.

ALGEBRAICAL ILLUSTRATION. It is required to find a point H in AB, or AB produced, such that

AB. BH=AH?. Let AB contain a units of length, and let AH contain x units ;

then BH=a - 2 : and x must be such that ala – x)= x?,

x2 + ax – - a?=0. Thus the construction for dividing a straight line in medial section corresponds to the solution of this quadratic equation, the two roots of which indicate the internal and external points of division.

or

EXERCISES. In the figure of 11. 11, shew that (i) if CH is produced to meet BF at L, CL is at right angles

to BF; (ii) if BE and CH meet at O, AO is at right angles to CH. (iii) the lines BG, DF, AK are parallel : (iv) CF is divided in medial section at A.

PROPOSITION 12. THEOREM.

In an obtuse-angled triangle, if a perpendicular is drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side on which, when produced, the perpendicular falls, and the line intercepted without the triangle, between the perpendicular and the obtuse angle.

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Let ABC be an obtuse-angled triangle, having the obtuse angle at C; and let AD be drawn from A perp. to the opp. side BC produced.

Then shall the sq. on AB be greater than the sum of the sqq. on BC, CA, by twice the rect. BC, CD.

Proof. Because BD is divided into two parts at C, .. the sq. on BD= the sum of the sqq. on BC, CD, with twice the rect. BC, CD.

II. 4, To each of these equals add the sq. on DA. Then the sqq. on BD, DA = the sum of the sqq. on BC, CD, DA, with twice the rect. BC, CD. But the sum of the sqq. on BD, DA= the sq. on AB, for the angle at D is a rt. angle.

I. 47. Similarly, the sum of the sqq. on CD, DA=the sq. on CA. ... the sq. on AB= the sum of the sqq. on BC, CA, with twice the rect. BC, CD.

That is, the sq. on AB is greater than the sum of the sqq. on BC, CA by twice the rect. BC, CD.

Q.E.D.

NOTE ON PROP. 12.

A general definition of the projection of one straight line on another is given on page 105. The student's attention is here called to a special case of projection which will enable us to simplify the Enunciation of Proposition 12.

A

4

P

D

In the above diagram, CA is a given straight line drawn from a point C in PQ; and from A a perpendicular AD is drawn to PQ. In this case, CD is said to be the projection of CA on PQ.

By applying this definition to the figure of Prop. 12, we see that the statement The sq. on AB is greater than the sum of the sqq. on BC, CA

by twice the rect. BC, CD is the particular form of the following general Enunciation :

In an obtuse-angled triangle the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of those sides, and the projection of the other side upon it.

. The Enunciation of Prop. 12 thus stated should be carefully compared with that of Prop. 13.

PROPOSITION 13. THEOREM.

In every triangle, the square on the side subtending an acute angle is less than the sum of the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle.

B

D
Fig. 1.

B
Fig. 2.

..

Let ABC be any triangle having the angle at C an acute angle; and let AD be the perp. drawn from A to the opp. side BC.

Then shall the sq. on AB be less than the sum of the sqq. on BC, CA, by twice the rect. BC, CD.

Proof. Now AD may fall within the triangle ABC, as in fig. 1, or without it, as in fig. 2. Because

sin fig. 1, BC is divided into two parts at D,

(in fig. 2, DC is divided into two parts at B,

in both cases the sum of the sqq. on BC, CD=twice the rect. BC, CD with the sq. on BD.

To each of these equals add the sq. on DA.
Then the sum of the sqq. on BC, CD, DA=twice the rect.

BC, CD with the sum of the sqq. on BD, DA.
But the sum of the sqq. on CD, DA= the sq. on CA,

for the angle ADC is a rt. angle. Similarly, the sum of the sqq. on BD, DA=the sq. on AB. ... the sum of the sqq. on BC, CA=twice the rect. BC, CD

with the sq. on AB. That is, the sq. on AB is less than the sum of the sqq. on BC, CA by twice the rect. BC, CD.

Q.E.D.

II. 7.

I. 47.

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