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In a right-angled triangle the square described on the hypotenuse is equal to the sum of the squares described on the other two sides.

H

K

B

LE

Let ABC be a right-angled triangle, having the angle BAC a right angle.

Then shall the square described on the hypotenuse BC be equal to the sum of the squares described on BA, AC.

Construction. On BC describe the square BDEC; I. 46. and on BA, AC describe the squares BAGF, ACKH. Through A draw AL parallel to BD or CE;

Proof

right angle,

and join AD, FC.

I. 31.

Then because each of the angles BAC, BAG is a

.. CA and AG are in the same straight line.

Now the angle CBD = the angle FBA,
for each of them is a right angle.

Add to each the angle ABC:

the whole angle FBC.

I. 14.

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.

But doubles of equals are equal:

I. 41.

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× 2 × 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.

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 BC: 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 perpendicular 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.

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 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 triangle BFD may be made to fill the space CKE.

Hence the two squares AF, HK

K

B

may be fitted together so as to H

G

form the single figure CBDE, which

F

will be found to be a perfect square, namely the square on the hypotenuse BC.

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.

D

B

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.

I. 3.

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,

the square on AD = the square on AB.

To each of these add the square on CA;

Constr.

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. .. the square on DC = the sum of the squares on CA, AD. I. 47. And, by hypothesis, the square on BC= the sum of the squares on CA, AB;

Because

.. the square on DC the square on BC:

also the side DC= the side BC.

Then in the triangles DAC, BAC,

DA = BA,

and AC is common to both;

Constr.

also the third side DC = the third side BC; Proved.

... the angle DAC = the angle BAC.

But DAC is a right angle.
Therefore also BAC is a right angle.

I. 8.

Constr.

Q.E.D.