and the square on BC is equal to the squares on BD, DC, and twice the rectangle BD, DC (11. 4);

therefore the square on BC is equal to the squares on AB, AC

(ax. 1):

wherefore the angle BAC is a right angle (1. 48). Q.ED.

10. If from the middle point of one of the sides of a rightangled triangle a perpendicular be drawn to the hypotenuse, the difference of the squares on the segments so formed is equal to the square on the other side.

Let ABC be a right-angled triangle, having the right angle BAC; and from D, the point of bisection of AC, let DE be drawn perpendicular to the hypotenuse BC.

Then the difference of the squares on BE, EC shall be equal to the square on AB.

Proof. The square on BD is equal to the squares on BA, AD (L. 47);

and the square on BD is also equal to the squares on BE, ED; therefore the squares on BA, AD are equal to the squares on BE, ED (ax. I):

but the square on AD, that is, CD (hyp.), is equal to the squares on ED, EC;

therefore the squares on AB, ED, EC are equal to the squares on BE, ED;

take from these equals the square on ED;

therefore the squares on AB, EC are equal to the square on BE (ax. 3);

again, take from these equals the square on EC;

wherefore the square on AB is equal to the difference of the squares on BE, EC. Q.E.D.

11. If A be the vertex of an isosceles triangle ABC, and CD be drawn perpendicular to AB, prove that the squares on the three sides are together equal to the square on BD, twice the square on AD, and thrice the square on CD.

B

Proof. The square on AC is equal to the squares on AD, CD (1. 47);

but AB is equal to AC (hyp.);

therefore the squares on AB, AC are equal to twice the squares on AD, CD:

the square on BC is equal to the squares on BD, CD; wherefore the squares on AB, AC, BC are together equal to

the square on BD, twice the square on AD, and thrice the square on CD (ax. 2). Q.E.D.

12. Enunciate Prop. 13 of the Second Book, and prove that the sum of the squares on any two sides of a triangle is equal to twice the sum of the squares on half the base, and of the line joining the vertical angle with the middle point of the base. (See deduction 4.)

13. If from any point perpendiculars be drawn to all the sides of any rectilineal figure, the sum of the squares on the alternate segments of the sides will be equal.

Let ABCDE be a rectilineal pentagon, and let perpendiculars be drawn from O to all the sides, viz. OF, OG, OH, OK, OL.

Then the sum of the squares on AF, BG, CH, DK, EL shall be equal to the sum of the squares on BF, CG, DH, EK, AL.

H

Join OA, OB, OC, OD, OE.

Proof-The square on OA is equal to the squares on AF, OF (1. 47);

OF:

and the square on OA is also equal to the squares on AL, OL: the square on OB is equal to the squares on BG, OG; and the square on OB is also equal to the squares on BF, the square on OC is equal to the squares on CH, OH; and the square on OC is also equal to the squares on CG, OG : the square on OD is equal to the squares on DK, OK; and the square on OD is also equal to the squares on DH, OH: the square on OE is equal to the squares on EL, OL; and the square on OE is also equal to the squares on EK, OK : therefore the sum of the squares on AF, OF, BG, OG, CH,

OH, DK, OK, EL, OL is equal to the sum of the squares on AL, OL, BF, OF, CG, OG, DH, OH, EK, OK (ax. 2); take from these equals the squares on OF, OG, OH, OK, OL; wherefore the sum of the squares on AF, BG, CH, DK, EL is equal to the sum of the squares on BF, CG, DH, EK, AL (ax. 3). Q.E.D.

14. If from the extremities of any chord in a circle lines be drawn to any point in the diameter to which it is parallel, the

Then the square on AB shall be equal to the squares on AC, CB, together with the rectangle AC, CB.

D

Draw AD perpendicular to BC produced (I. 12); produce CD to E, making DE equal to DC (1. 3); and join AE (post. 1).

Proof. In the triangles ACD, AED,

the two sides CD, DA are equal to the two sides ED, DA, each to each,

and the angle CDA is equal to the angle EDA (ax. 11);
therefore the base AC is equal to the base AE (1. 4),
and the angle ACD is equal to the angle AED.

The angle ACB is two-thirds of two right angles (hyp.); therefore the angle ACE is one-third of two right angles (1. 13): wherefore the angle AEC is a third of two right angles (ax. 1), and the angle CAE is also a third of two right angles (1. 32): therefore the triangle ACE is equiangular;

wherefore the triangle ACE is equilateral (1. 6, Cor.).

The square on AB is equal to the squares on AC, CB, together with twice the rectangle BC, CD (II. 12);

but CD is half of CE, and therefore half of CA;

wherefore the square on AB is equal to the squares on AC, CB, together with the rectangle AC, CB. Q.E.D.

DEDUCTIONS.

1. If a straight line be divided into two equal, and also into two unequal parts, the part between the points of section is equal to half the difference of the unequal parts.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D. Then CD shall be equal to half the difference of AD, DB.

Proof-AD is equal to the sum of AC, CD;
DB is equal to the difference of CB, CD,

ind therefore DB, CD are together equal to CB (ax. 2);
out CB is equal to AC (hyp.);

herefore AD is equal to DB, and twice CD;

ake from these equals DB;

herefore the difference of AD, DB is equal to twice CD (ax. 3); wherefore CD is equal to half the difference of AD, DB. Q.E.D.

2. If the sides of a triangle be as 2, 4, 5, show whether it will be acute or obtuse angled.

As the square of the longest side is greater than the sum of the squares on the other two sides, the triangle is obtuseingled (11, 12). Q.E.D.

3. Divide a given straight line into two parts, so that the square on one part shall be double the square on the other.

Let AB be the given straight line.

It is required to divide AB into two parts, so that the square on one of the parts shall be double the square on the other part.

D