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Then in the triangles ABD, FBC,

AB= FB, Because

and BD=BC, | also the angle ABD= the angle FBC; Proved. .. the triangle ABD = the triangle FBC.

I. 4. Now the parallelogram BL is double of the triangle ABD, being on the same base BD, and between the same parallels BD, AL.

I. 41. And the square GB is double of the triangle FBC, being on the same base FB, and between the same parallels FB, GC.

1. 41. But doubles of equals are equal : Ax. 6. therefore the parallelogram BL= the square GB. Similarly, by joining AE, BK it can be shewn that the parallelogram CL = the square CH. Therefore the whole square Be=the sum of the squares

GB, HC: that is, the square described on the hypotenuse BC is equal

to the sum of the squares described on the two sides BA, AC.

Q.E.D.

Note. It is not necessary to the proof of this Proposition that the three squares should be described external to the triangle ABC; and since each square may be drawn either towards or away

from the triangle, it may be shewn that there are 2 x 2 x 2, or eight, possible constructions.

Obs. The following properties of a square, though not formally enunciated by Euclid, are employed in subsequent proofs. [See I. 48.]

(i) The squares on equal straight lines are equal.
(ii) Equal squares stand upon equal straight lines.

EXERCISES ON PROPOSITION 47.

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1. In the figure of this Proposition, shew that

(i) If BG, CH are joined, these straight lines are parallel ; (ii) The points F, A, K are in one straight line ; (iii) FC and AD are at right angles to one another ; (iv) If GH, KE, FD are joined, the triangle GAH is equal

to the given triangle in all respects; and the triangles FBD, KCE are each equal in area to the triangle ABC.

[See Ex. 9, p. 79.] 2. On the sides AB, AC of any triangle ABC, squares ABFG, ACKH are described both toward the triangle, or both on the side remote from it: shew that the straight lines BH and CG are equal.

3. On the sides of any triangle ABC, equilateral triangles BCX, CAY, ABZ are described, all externally, or all towards the triangle : shew that AX, BY, CZ are all equal.

4. The square described on the diagonal of a given square, is double of the given square.

5. ABC is an equilateral triangle, and AX is the perpendicular drawn from A to B : shew that the square on AX is three times the square on BX.

6. Describe a square equal to the sum of two given squares.

7. From the vertex A of a triangle ABC, AX is drawn perpendi. cular to the base : shew that the difference of the squares on the sides AB and AC, is equal to the difference of the squares on BX and CX, the segments of the base.

8. If from any point O within a triangle ABC, perpendiculars OX, OY, OZ are drawn to the sides BC, CA, AB respectively : shew that the sum of the squares on the segments AZ, BX, CY is equal to the sum of the squares on the segments AY, CX, BZ.

9. ABC is a triangle right-angled at A; and the sides AB, AC are intersected by a straight line PQ, and BQ, PC are joined. Prove that the sum of the squares on BQ, PC is equal to the sum of the squares on BC, PQ.

10. In a right-angled triangle four times the sum of the squares on the two medians drawn from the acute angles is equal to five times the square on the hypotenuse.

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NOTES ON PROPOSITION 47.

It is believed that Proposition 47 is due to Pythagoras, a Greek philosopher and mathematician, who lived about two centuries before Euclid.

Many experimental proofs of this theorem have been given by means of actual dissection : that is to say, it has been shewn how the squares on the sides containing the right angle may be cut up into pieces which, when fitted together in other positions, exactly make up the square on the hypotenuse. Two of these methods of dissection are given below.

I. In the adjoining diagram ABC is the given right-angled triangle, and the figures AF, HK are the squares on AB, AC, placed side by side.

FD is made equal to EH or AC; and the two squares AF, HK are cut

А.

B along the lines ED, DB.

Then it will be found that the triangle EHD may be placed so as to fill up the space CAB; and the

EA

K triangle BFD may be made to fill the space CKE.

Hence the two squares AF, HK may be fitted together so

H G

D F form the single figure CBDE, which will be found to be a perfect square, namely the square on the hypotenuse BC.

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PROPOSITION 48. THEOREM. If the square described on one side of a triangle be equal to the sum of the squares described on the other two sides, then the angle contained by these two sides shall be a right angle.

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I. 3.

Let ABC be a triangle ; and let the square described on BC be equal to the sum of the squares described on BA, AC.

Then shall the angle BAC be a right angle. Construction. From A draw AD at right angles to AC; I. 11.

and make AD equal to AB.

Join DC. Proof. Then, because AD=AB, Constr. the square on AD= the square on AB.

To each of these add the square on CA ; then the sum of the squares on CA, AD = the sum of the squares on CA, AB.

But, because the angle DAC is a right angle, Constr.

square on DC = the sum of the squares on CA, AD. 1.47. And, by hypothesis, the square on BC = the sum of the squares on CA, AB;

.. the square on DC = the square on BC:
... also the side DC = the side BC.
Then in the triangles DAC, BAC,
DA= BA,

Constr. Because and AC is common to both;

also the third side DC = the third side BC; Proved, .. the angle DAC = the angle BAC.

I. 8. But DAC is a right angle.

Constr, Therefore aļso BAC is a right angle. Q.E.D.

... the

INTRODUCTORY.

HINTS TOWARDS THE SOLUTION OF GEOMETRICAL EXERCISES.

ANALYSIS.

SYNTHESIS.

It is commonly found that exercises in Pure Geometry present to a beginner far more difficulty than examples in any other branch of Elementary Mathematics. This seems to be due to the following

causes :

(i) The variety of such exercises is practically unlimited; and it is impossible to lay down for their treatment any definite methods, such for example as the rules of Elementary Arithmetic and Algebra.

(ii) The arrangement of Euclid's Propositions, though perhaps the most convincing of all forms of argument, affords in most cases little clue as to the way in which the proof or construction was discovered.

Euclid's propositions are arranged synthetically: that is to say, starting from the hypothesis or data, they first give a construction in accordance with postulates, and problems already solved; then by successive steps based on known theorems, they prove what was required in the enunciation.

Thus Geometrical Synthesis is a building up of known results, in order to obtain a new result.

But as this is not the way in which constructions or proofs are usually discovered, we draw the student's attention to the following hints.

Begin by assuming the result it is desired to establish ; then by working backwards, trace the consequences of the assumption, and try to ascertain its dependence on some simpler theorem which is already known to be true, or on some condition which suggests the necessary construction. If this attempt is successful, the steps of the argument may in general be re-arranged in reverse order, and the construction and proof presented in a synthetic form.

This unravelling of a proposition in order to trace it back to some earlier principle on which it depends, is called geometrical analysis : it is the natural way of attacking many theorems, and it is especially useful in solving problems.

Although the above directions do not amount to a method, they often furnish a mode of searching for a suggestion. Geometrical Analysis however can only be used with success when a thorough grasp of the chief propositions of Euclid has been gained.

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