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[This proposition, which is deduced directly from props. 12, 13, is not given by Euclid, but it seems to deserve a place in the second Book, not only on account of its own interest, but because also it is the key to a large class of problems, which involve by the enunciations the bisection of some line. Some examples of such problems will now be given.]

1. If straight lines be drawn from each angle of a triangle to bisect the opposite sides, four times the sum of the squares on these lines is equal to three times the sum of the squares on the sides.

2. If AB, one of the equal sides of an isosceles triangle ABC, be produced beyond the base to D, so that BD = AB, show that the square on CD is equal to the square on AB together with twice the square on BC.

3. From any point P lines are drawn to the angles of a rectangle ABCD; show that the squares on PA, PC are together equal to the squares on PB, PD.

4. A square BDEC is described on the hypotenuse of a right-angled triangle ABC ; show that the squares on DA, AC are together equal to the squares on EA, AB.

5. The squares on the diagonals of a parallelogram are together equal to the squares on the four sides.

6. If two points be taken in the diameter of a circle equally distant from the centre, the sum of the squares of two lines drawn from these points to any point in the circumference will be constant.

7. The hypotenuse of a right-angled triangle ABC is trisected at D, E, and CD, CE are joined ; show that the sum of the squares on the sides of the triangle CDE is equal to two-thirds of the square on AB.

8. In any quadrilateral, the squares on the diagonals are together double of the sum of the squares on the two lines joining the bisections of the opposite sides.

G

PROPOSITION XIV. PROBLEM.

To describe a square that shall be equal to a given rectilineal figure.

Let A be the given rectil. figure.

It is required to describe a square that shall be equal to A.

[blocks in formation]

EF =

Constr. Describe the rectangular O BCDE = the rectil. figure A.

(i. 45) Then if BE = ED, it is a sq., and what was required is

done. But if BE be not = ED, produce BE to F, and make

ED, bisect BF in G, (i. 10) from centre G, at distance GB or GF, describe semicircle BHF, and produce DE to meet the circumference in H.

The sq. on EH shall equal the given figure A. Dem. :: BF is divided into two equal parts at G and

into two unequal parts at E, .-, rect. BE, EF with sq. on EG = sq. on GF;

(ii. 5) but GF=GH, and .. sq. on GF = sq. on GH, and .. = sqs. on GE, EH ;

(i. 47) :, rect. BE, EF, with sq. on EG = sqs. on EG, EH ;

take away from each the sq. on EG ;

and rect. BE, EF = sq. on EH.

But rect. BE, EF is BD : EF = ED, (constr.) .: BD= sq. on EH ; and BD = A; (constr.) :: sq. on EH

rectil. fig. A. Therefore, a square has been described as required.

Q. E. F.

[blocks in formation]

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