Show that the sum of the squares on two lines is never less than twice their rectangle.

Let the straight line AB be divided into any two parts in the point C.

Then the sum of the squares on AB, BC shall not be less than twice the rectangle AB, BC.

Proof-The squares on AB, BC are equal to twice the rectangle AB, BC, together with the square on AC (11. 7); that is, the sum of the squares on two lines is equal to twice their rectangle, and the square on the difference of the two lines:

wherefore the sum of the squares on two lines is never less than twice their rectangle. Q.E.D.

Note.-If the two lines be equal, the sum of their squares is equal to twice their rectangle.

9. Divide a straight line into two parts so that the sum of their squares may be the least possible.

Let AB be the given straight line.

It is required to divide it into two parts so that the sum of their squares may be the least possible.

A

C

D

B

Bisect AB in C (I. 10).

Then the sum of the squares on AC, CB shall be the least possible.

Proof. Take any other point D in AB.

The squares on AD, DB are equal to twice the squares on CD, CB (11. 9);

therefore twice the square on CB, that is, the sum of the squares on AC, CB, is less than the squares on AD, DB.

Wherefore the sum of the squares on AC, CB is the least possible. Q.E.D.

Note. This deduction is the same as that to II. 4.

11. Show that the squares on the whole line and one part are equal to three times the square on the other part.

Let AB be divided in the point H, so that the rectangle AB, BH is equal to the square on AH.

Then the squares on AB, BH shall be equal to three times the square on AH.

Proof. The rectangle AB, BH is equal to the square on AH (hyp.);

therefore twice the rectangle AB, BH is equal to twice the square on AH:

to each of these equals add the square on AH;

therefore twice the rectangle AB, BH, together with the square

on AH, is equal to three times the square on AH (ax. 2); but the squares on AB, BH are equal to twice the rectangle AB, BH, together with the square on AH (II. 7); wherefore the squares on AB, BH are equal to three times the square on AH (ax. 1). Q.E.D.

12. ABDE is a square on AB, the hypotenuse of a rightangled triangle ABC. Show that the difference of the squares

on CD, CE is equal to the difference of the squares on CB, CA.

G

B

Through C draw FCG parallel to AB or ED (1. 31), and produce EA, DB to meet FCG in F, G (post. 2).

Proof. The angles at F, G are right angles (1. 29).

The angle CBD is obtuse, being greater than the right angle ABD;

and the angle CAE is obtuse, being greater than the right angle BAE.

The square on CD is equal to the squares on CB, BD, and twice the rectangle DB, BG (II. 12);

and the square on CE is equal to the squares on CA, AE, and twice the rectangle EA, AF;

but BD is equal to AE (hyp.), and BG is equal to AF (1. 34) ; wherefore the difference of the squares on CD, CE is equal to the difference of the squares on CB, CA (ax. 3). Q.E.D.

If the obtuse angle be two-thirds of two right angles, show that the square of the opposite side is equal to the squares of the containing sides, together with the rectangle of those sides.

Let ABC be an obtuse-angled triangle, having the obtuse angle ACB equal to two-thirds of two right angles.

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 (1. 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.

21. Divide a right angle into three equal parts.

Let ABC be the given right angle.
It is required to trisect it.

B

Upon AB construct the equilateral triangle ABD (1. 1); bisect the angle ABD by the straight line BE (1. 9). Then BE, BD shall trisect the right angle ABC.

Proof. Because the triangle ABD is equilateral,

therefore the triangle ABD is equiangular (1. 5, Cor.)

the three angles of a triangle are together equal to two right angles (1. 32),

therefore the angle ABD is equal to a third of two right angles, that is, two-thirds of a right angle;

wherefore each of the angles ABE, EBD is a third of a right angle;

and because the angle ABD is two-thirds of a right angle, therefore the angle DBC is a third of a right angle.

Wherefore the right angle ABC is trisected by BE, BD. Q.E.F.

22. Show that if two lines bisect the opposite sides of a trapezium, they also bisect each other.

Let ABCD be a trapezium, and let E, F, G, H be the points of bisection of the sides.