equal to the square of EB; And the squares of BA, AE are equal (47. 1.) to the square of EB, because the angle EAB is a right angle; therefore the rectangle CF.FA, together with the square of AE, is equal to the squares of BA, AE: take away the square of AE, which is common to both, therefore the remaining rectangle CF.FA is equal to the square of AB. Now the figure FK is the rectangle CF.FA, for AF is equal to FG; and AD is the square of AB; therefore FK is equal to AD: take away the common part AK, and the remainder FH is equal to the remainder HD. But HD is the rectangle AB.BH for AB is equal to BD; and FH is the square of AH; therefore the rectangle AB.BH is equal to the square of AH: Wherefore the straight line AB is divided in H, so that the rectangle AB.BH is equal to the square of AH.

PROP. XII. THEOR.

In obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted between the perpendicular and the obtuse angle.

Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn (12. 1.) perpendicular to BC produced: The square of AB is greater than the squares of AC, CB, by twice the rectangle BC.CD.

=

Because the straight line BD is divided. into two parts in the point C, BD2=(4. 2.) BC2+CD2+2BC.CD; add AD to both: Then BD2+AD2 BC2+ CD2+ AD2+ 2BC.CD. But AB2-BD2+AD2 (47. 1.), and AC2= CD2+ AD2 (47. 1.); therefore, AB2=BC2+AC2+2BC.CD; that is, AB2 is greater than BC2+AC2 by 2BC.CD.

PROP. XIII. THEOR.

In every triangle the square of the side subtending any of the acute angles, is less than the squares of the sides containing that angle, by twice the rectan gle contained by either of these sides, and the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle.

Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular (12. 1.) AD from the opposite angle: The square of AC, opposite to the angle B, is less than the squares of CB, BA by twice the rectangle CB.BD.

First, let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D (7. 2.), BC2+ BD2-2BC. BD+CD2. Add to each AD2; then BC2+BD2+AD2=2BC.BD+CD2+ AD2. But BD2+AD2=AB2, and CD2+ DA2 AC2 (47. 1.); therefore BC2+AB2= 2BC.BD+AC2; that is, AC2 is less than BC2+AB2 by 2BC.BD.

A

D

C

B Secondly, let AD fall without the triangle ABC:* Then because the angle at D is a right angle, the angle ACB is greater (16. 1.) than a right angle, and AB2 (12. 2.) AC2+BC2+2BC.CD. Add BC2 to each; then AB2+BC2=AC2+2BC2+2BC.CD. But because BD is divided into two parts in C, BC2+BC.CD=(3. 2.) BC.BD, and 2BC2+2BC.CD =2BC.BD: therefore AB2+ BC2=AC2+ 2BC.BD; and AC2 is less than AB2+BC2, by 2BD.BC.

Lastly, let the side AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifest that (47. 1.) AB2+BC2= AC2+2BC2 AC2+2BC.BC.

A

B

C

PROP. XIV. PROB.

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

Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A.

Describe (45. 1.) the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it, BE, ED are equal to one another, it is a square, and what was required is done; but if they are not equal, produce one of them, BE to F, and make EF equal to ED, and bisect BF in G; and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H, and join GH. Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the rectangle BE.EF, together with the square of EG, is equal (5. 2.) to the square of GF: but GF is equal to GH; therefore the rectangle BE, EF, together with the square of EG, is equal to the square of GH: But the squares of

See figure of the last Proposition

HE and EG are equal (47. 1.) to the square of GH: Therefore also the rectangle BE.EF, together with the square of EG, is equal to the squares of HE and EG. Take away the square of EG, which is common to both, and the remaining rectangle BE.EF is equal to the square of EH: But BD is the rectangle con

tained by BE and EF, because EF is equal to ED; therefore BD is equal to the square of EH; and BD is also equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH.

PROP. A. THEOR.

If one side of a triangle be bisected, the sum of the squares of the other two sides is double of the square of half the side bisected, and of the square of the line drawn from the point of bisection to the opposite angle of the triangle.

Let ABC be a triangle, of which the side BC is bisected in D, and DA drawn to the opposite angle; the squares of BA and AC are together double of the squares of BD and DA.

A

From A draw AE perpendicular to BC, and because BEA is a right angle, AB2=(47. 1.) BE2+AE2 and AC2= CE2+AE; wherefore AB2+AC2=BE2 +CE2+2AE2. But because the line BC is cut equally in D, and unequally in E, BE2+ CE2 (9. 2.) 2BD2+ 2DE2; therefore AB2 + AC2=2BD2 + 2DE2.2AE2.

=

Now DE2+AE2=(47. 1.) AD2, and 2DE2+2AE2-2AD2; wherefore AB2+ AC2-2BD2+2AD2.

B

D

E

PROP. B. THEOR.

The sum of the squares of the diameters of any parallelogram is equal to the sum of the squares of the sides of the parallelogram.

Let ABCD be a parallelogram, of which the diameters are AC and BD ; the sum of the squares of AC and BD is equal to the sum of the squares of AB, BC, CD, DA.

Let AC and BD intersect one another in E: and because the vertical angles AED, CEB are equal (15. 1.), and also the alternate angles EAD,

ECB (29. 1.), the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each; but the sides AD and BC, which are opposite to equal angles in these triangles, are also equal (34. 1.); therefore the other sides which are opposite to the equal angles are also equal (26. 1.), viz. AE to EC, and ED to EB.

B

A

D

E

C

Since, therefore, BD is bisected in E, AB2+AD2=(A. 2.) 2BE2+2AE2; and for the same reason, CD2+ BC2 = 2BE2+2EC2-2BE2+2AE2, because EC-AE. Therefore AB2+AD2 +DC2+BC2=4BE2+4AE2. But 4BE2=BD2, and 4AE2=AC2 (2. Cor. 8. 2.) because BD and AC are both bisected in E; therefore AB2+ AD2+CD2+BC2=BD2+AC2.

COR. From this demonstration, it is manifest that the diameters of every parallelogram bisect one another.

SCHOLIUM.

In the case of the rhombus, the sides AB, BC, being equal, the triangles BEC, DEC, have all the sides of the one equal to the corresponding sides of the other, and are therefore equal: whence it follows that the angles BEC, DEC, are equal; and, therefore, that the two diagonals of a rhombus cut each other at right angles.

ELEMENTS

OF

GEOMETRY.

BOOK III.

DEFINITIONS.

A. The radius of a circle is the straight line drawn from the centre to the circumference.

1. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it.

And that line which has but one point in common with the circumference, is called a tangent, and the point in common, the point of contact.

2. Circles are said to touch one another, which meet, but do not cut one another.

3. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal.

4. And the straight line on which the greater perpendicular falls, is said to be farther from the centre.

B. Any portion of the circumference is called an arc.

The chord or subtense of an arc is the straight line which joins its two extremities.

C. A straight line is said to be inscribed in a circle, when the extremities of it are in the circumference of the circle. And any straight line which meets the circle in two points, is called a secant.

5. A segment of a circle is the figure contained by a straight line, and the arc which it cuts off.