(1) BAC is a right-angled triangle and squares are described on its three sides. The parts a, b, c, d, e, out of the two smaller squares will be found to just fit together into the larger square.

(2) Take the two squares PQRS and ARMC. Place them together and then cut off the parts 1 and 2. It will be found that these two parts fit into the parts of the two squares left, and form the square PBCD.

(3) Take o, the centre of the square FBAG and draw the lines mon and kol, respectively perpendicular and parallel to BC.

Take p, q, r, s, the middle points of CB, BD, DE, and EC; and through p, q, r, s, draw lines parallel to AC and AB alternately.

It will be found that the square tuvw is equal to CH and that the four parts of BG make up the square BE.

(4) ABCD and EFGH are two equal squares.

Take any point k in AD, and mark off Bl, Cm, Dn, Ep, Er, all equal to Ak
Join kl, lm, mn, nk; and through p and r draw ptq and rts parallel to
EF and EH respectively.
Join Ft, Ht.

Proof.—(i) By Euclid I. 4 we may prove the triangles Akl, Blm, Cmn,
Dnk, Frt, Fqt, Hpt, and Hts all equal.

(ii) The figures klmn, rp, and qs may be proved to be squares.
Because ABCD is equal to EFGH,

from each take four equal triangles;

therefore the remainder, the square klmn, is equal to the
remainders, the squares rp and qs.

But klmn is the square on the hypotenuse of one of the equal triangles,
and rp and qs are equal to the squares on the two sides of
one of the equal triangles;

therefore the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Q. E. D.

(5) In the figure of Prop. 47 describe a triangle DME on DE on the side
remote from A, and make DM equal to AC, and ME equal
to BA; join GH, FAK, and AM.

Then the quadrilaterals FGHK, FBCK, ABDM, and ACEM may be
proved equal to one another;

therefore the square on BC is equal to the sum of the squares on BA, AC.

Q. E. D.

Many other proofs of Prop. 47 have been given. Of course any of the proofs (1), (2), (3) may be completed by proving the equality of the corresponding parts of the squares.

EXERCISES ON PROPOSITIONS 47 AND 48.

1. In the figure of Prop. 47 :

(a) If AL cuts BC in M, show that the square on BA is equal to the rectangle contained by BC and BM.

(b) If FC and AD intersect in N, prove that the angle ANF is equal to the right angle FBA.

(c) Join GH, FD, KE and prove that the triangles GAH, FBD, and KCE are each equal to ABC.

(d) Join BK and show that FC, AL, BK all pass through the same point. (e) Join BG and CH, and show that these lines are parallel.

2. The sum of the squares on the sides of a rhombus is equal to the sum of the squares on its diagonals.

3. In an equilateral triangle the sum of the squares on the three sides is equal to four times the square on the perpendicular from the vertex to the base. 4. If two right-angled triangles have their hypotenuses equal, and have also one side equal to one side, the two triangles shall be equal in all respects. 5. ABC is any triangle, and AD is drawn perpendicular to BC. Show that the difference of the squares on AB, AC is equal to the difference of the squares on BD, CD.

6. Prove that if the difference of the squares on two sides of a triangle is equal to the square on the third side, the triangle is right-angled.

7. In Prop. 47 show that F, A, K are in one and the same straight line. 8. Show how to describe a square equal to the sum of two given squares.

NOTES ON PROPOSITIONS 33-48.

Proposition 33.-This is an easy rider on Prop. 4.

If AB and CD are joined, but not towards the same parts, then AD and BC bisect each other.

Proposition 34.-This is a rider on Prop. 26.

In the third part of the proof care should be taken to mention first the two parts of the same angle as e. g. of BAD.

Euclid proves the triangles DAC, BAC equal in area by Prop. 4; this was necessary, because in Prop. 26 Euclid originally said nothing about the area of the triangles.

Proposition 35.—The second and important part of the proof in Prop. 35 depends on proving the triangles FDC, EAB equal in area by Prop. 4.

In works on Mensuration the area of a rectangle is said to be equal to the product of the length by the breadth of the rectangle. By Prop. 35 the area of a parallelogram is equal to the area of a rectangle on the same base and of equal altitude. It follows therefore that the area of any parallelogram is equal to the product of the base into the altitude of the parallelogram.

Proposition 36.--This is an easy rider on Prop. 35 and Axiom 1. Propositions 37 and 38.-These are riders on Props. 35 and 36 and Axiom 7. Propositions 39 and 40.-These are proved by Props. 37 and 38 and Reductio ad absurdum.

Proposition 41.-This is a rider on Props. 37 and 34.

Since the area of any parallelogram is equal to the product of the base into the altitude of the parallelogram, it follows from Prop. 41 that the area of a triangle is equal to half the product of the base into the altitude of the triangle.

Additional Proposition. This should logically precede Prop. 42. It depends on Prop. 37. By the aid of this Additional Proposition we can see how to bisect a quadrilateral figure. (1) If the bisecting line must pass through an angle, as D; bisect EC at X; then the triangle DXC is half DEC, and therefore half ABCD. (2) If the bisecting line must pass through a certain point P in BC; as before draw DX to the middle of EC; join DP and draw XQ parallel to DP, meeting CD in Q. Then PQC is equal to DXC, and is therefore half of ABCD.

Proposition 42.—The proof of this problem depends on Prop. 41.

With this proposition Euclid begins a series of problems dealing with the construction of parallelograms. Before doing this he ought perhaps to have dealt with the construction of triangles. In Prop. 22 we are shown how to describe a triangle having its sides equal to three given straight lines. If therefore we are required to describe a triangle equal to a given triangle, we can describe a triangle having its sides equal, each to each, to the three sides of the given triangle, and the two triangles will be equal in all respects.

But Euclid has given no proposition showing how to describe a triangle equal to a given rectilineal figure. A solution of this problem will be found on page 83.

In the construction of parallelograms four cases occur:

(1) To describe a parallelogram equal to a given triangle and having an angle equal to a given angle. This is solved in Prop. 42.

(2) On a given straight line to describe a parallelogram equal to a given triangle, and having an angle equal to a given angle. This is solved in Prop. 44.

(3) To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given angle. This is solved for a quadrilateral figure in Prop. 45, and the solution for any rectilineal figure is similar.

(4) On a given straight line to describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given angle. This is not given in Euclid, because the method of solution is like that given in Prop. 44.

EUC.

H

Proposition 43.-This is an easy rider on Prop. 34, and it is often used.

Proposition 44.—This depends on Prop. 43. Notice that the hardest part of the proposition is taken up with proving that the construction is possible, namely that FE and HB will meet if produced.

Proposition 45.-This depends on Props. 42 and 44. Notice again that the hardest part of the proposition is taken up with proving that FGLM is a parallelogram, inasmuch as GHL and FKM are straight lines. If the angles

FGH and KHL were not equal, GHL need not be a straight line.

Proposition 46.-The proof of this proposition falls into two parts. In the first part we prove the sides all equal by Prop. 34; and in the second part we prove the angles all right angles by Props. 29 and 34.

Proposition 47.-The proof of this proposition falls into four parts as follows:

(1) GA and AC are in one straight line (by Prop. 14).

(2) The angle DBA is equal to the angle FBC.

(3) The triangle DBA is equal to the triangle FBC in area (by Prop. 4).

(4) BL is equal to BG (by Prop. 41 and Axiom 6).

This theorem is said to have been discovered by Pythagoras, who lived in the south of Italy between 570 and 500 B.C. It is often called the Theorem of Pythagoras. It is so very important that we give several alternative proofs of it, and also a number of Exercises and Riders on it afterwards.

Proposition 48.-The proof of this proposition falls into two parts, viz.: (1) BC is proved equal to DC by Prop. 47, and (2) the angle BAC is proved equal to the angle DAC by Prop. 8.

It is very important to remember that in the construction we make CAD a right angle; we do not produce BA to D. In order to remind the student of this, the letter R is placed in the angle CAD.

In proving this proposition Euclid makes use of the two following theorems, although he has not proved them before:

(1) The squares described on equal straight lines are equal.

(2) If two squares are equal, the lines on which they are described are equal.