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make AD equal to AB;
through D draw DE parallel to AB,
and through B draw BE parallel to AD.

Then ABED shall be the square required.

(I. 3) (I. 31)

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B

A
Now ABED is a parallelogram;

(Constr.) therefore AB is equal to DE, and AD to BE; (I. 34) but AB is equal to AD;

(Constr.) therefore BA, AD, DE, EB are all equal to one another,

and the parallelogram ADEB is equilateral.

Again, since AD meets the parallels AB, DE, the angles BAD, ADE are equal to two right angles; (I. 29) but BAD is a right angle;

(Constr.) therefore also ADE is a right angle ; therefore each of the opposite angles ABE, BED is a right angle,

(I. 34) wherefore the figure ADEB is rectangular. And it has been proved to be equilateral; therefore it is a square.

(Def. 30) Wherefore a square ADEB has been described upon the given straight line AB.

Q. E. F. CoR.-Every parallelogram that has one right angle has all its angles right angles.

102. EXERCISE LIV. 1. Where is the proposition found in Euclid—that, if a parallelogram has one angle a right angle, it is a rectangle or oblong?

2. If a parallelogram were constructed with two adjacent sides equal, and one angle right, what would it become ?

What proposition has its construction identical with this ?

3. Draw a figure like that of XLVI., and join CB, CE; then prove that the triangle CBE is half the square AE.

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B В

4. Given, in the adjacent figure, BAC a right angle and CH a square ; what is the sum of the adjacent angles at A?

5. Draw an inference by I. 14.

6. Given, also, that BE is a square; prove the angles BCE, ACK equal.

7. Thence prove angles ACE, BCK equal.

8. Also show that EC, CA are equal to BC, CK, each to each.

9. From the results of 7 and 8, find two triangles equal by I. 4.

10. Prove, in the same figure, BH and CK are parallel.

11. Given also that AL is parallel to CE, find parallelograms respectively double of the triangles ACE and BCK.

12. Draw an inference by Ax. 6.

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103, PROPOSITION XLVII.—THEOREM.

In any right-angled triangle, the square which is described upon the side subtending the right angle is equal to the squares described upon the sides which contain the right angle.

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

upon

the side BC shall be equal to the squares

described

upon BA, AC. On BC describe the square BDEC,

(I. 46) and on BA, AC the squares GB, HC; through A draw AL parallel to BD or CE;

(I. 31) and join AD, FC.

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Then because the angle BAC is a right angle, (Hyp.) and that BAG is also a right angle,

(Def. 30) the two straight lines AC, AG, upon the opposite sides of AB,

make with it at the point A the adjacent angles together

equal to two right angles ; therefore CA is in the same straight line with AG. (I. 14) For the same reason, BA and AH are in one straight line.

Now the angle DBC is equal to the angle FBA, each of them being a right angle; add to each the angle ABC;

then the whole angle DBA is equal to the whole FBC. (Ax. 2) And because the two sides AB, BD are equal to the two

FB, BC, each to each, and the included angle ABD is equal to the includedangle FBC;

therefore the triangle ABD is equal to the triangle FBC. (1.4)

Now the parallelogram BL is double of the triangle ABD(I.41) because they are upon the same base BD, and between the

same parallels BD, AL; also the square GB is double of the triangle FBC, because they are upon the same base FB, and between the

same parallels FB, GC; therefore BL is equal to the square BG.

.

(Ax. 6) Similarly, by joining AE, BK, it can be proved

that CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB, HC ;

(Ax. 2) that is, the square upon the side BC is equal to the squares

upon the sides AB, AC. Therefore, in any right-angled triangle, &c.

Q. E. D. 104, EXERCISE LV. 1. Give fully the proof showing that the parallelogram CL and the square CH are equal.

2. Prove that the square upon AD, in the figure to XLVII., is equal to the squares on AL, LD.

3. If CD be joined in the same figure, the square upon CD is equal to the squares upon AB, BC, CA.

4. In any right-angled triangle ABC, having C the right angle, the square upon BC is equal to the difference between the squares on AB and AC.

5. The square described upon the diagonal of a square is equal to twice the latter square.

6. In a rectangle the sum of the squares on the two diagonals is equal to the sum of the squares on the sides.

Note.—Proposition XLVII. is often called the Theorem of Pythagoras, from the name of its discoverer, an ancient Greek philosopher, who lived in the sixth century before Jesus Christ.

It has important uses in arithmetical calculations, as well as in Geometry. If the learner could practice some of these, while studying this part of Geometry, it would be an advantage.

105. PROPOSITION XLVIII.—THEOREM. square

described upon one of the sides of a triangle be equal to the

squares
described

upon the other two sides of it, the angle contained by these two sides is a right angle. Let the square described upon BC, one of the sides of the

triangle ABC, be equal to the squares upon the other two sides AB, AC. Then the angle BAC shall be a right angle.

D

If the

B

From the point A draw AD at right angles to AC; (I. 11) make AD equal to AB, and join DC.

Then, because AD is equal to AB, the square upon AD is equal to the square upon AB ; to each of these equals add the square on AC; then the squares on AD, AC are equal to the squares on AB, AC. But the squares on AD, AC are equal to the square on DC,(I. 47) because DAC is a right angle; and the squareson BA, ACare equal to the square on BC; (Hyp.)

therefore the square on DC is equal to the square on BC;

wherefore the side DC is equal to the side BC.

And because AD is equal to AB, and AC is common to the triangles DAC, BAC; therefore DA, AC are equal to BA, AC, each to each ; and the base DC has been proved equal to the base BC; therefore the angle DAC is equal to BAC.

(I. 8) But DAC is a right angle,

(Constr.) therefore also BAC is a right angle, Therefore, if the square, &c.

Q. E. D.

106. EXERCISE LVI.

1. State the relation of XLVIII. to XLVII.

2. Is the square on one side of an isosceles triangle be half the square on another, prove the triangle is also right-angled.

3. Given two squares, show how to find a square equal in area to their

sum.

4. Given a square, show how to find a square which is its double in area. 5. Given a square, find another of three times its area.

6. Given three squares, find another whose area is equal to the sum of the three.

7. Show that the square on the diagonal of a cube is three times one face of the cube. 8. State the proposition of which the enunciation may be thus expressed :

“ Construct an equilateral rectangle, each side of which shall be

equal to a given straight line." 9. Give the proposition equivalent to :

The square on the hypotenuse of a right-angled triangle is equal to the sum of the squares upon the other two sides.” 10. In the construction to XLVIII., show that it would be unsatisfactory to say, instead of the first line,

“ Produce B A to D,unless we choose to add a short piece of reasoning to justify it.

Give the reasoning which would be required.

11. State the assumption made in the foregoing proof where the inference is drawn:

· Wherefore the side DC is equal to the side BC."

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