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 without the triangle 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.

A

Because the straight line BD is divided into two parts in the point C, the square of BD is equal (4. 2.) to the squares of BC, CD, and twice the rectangle BC, CD: to each of these equals add the square of DA; and the squares of BD DA are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD: but the square of BA is equal (47. 1.) to the squares of BB, DA, because the angle at D is a right angle; and the square of CA is equal (47. 1.) to the squares of CD, DA: therefore the square of BA is equal to the squares of BC, CA, and twice the rectangle BC, CD; that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BC, CD. Therefore, in obtuse angled triangles, &c. Q. E. D.

B

PROP. XIII. THEOR.

C

D

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 rectangle 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 BBC 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.

*See Note..

A

First, let AD fall within the triangle ABC; and because the straight line CD is divided into two parts in the point D, the squares of CB, BD are equal (7. 2.) to twice the rectangle contained by CB, BD, and the square of DC: to each of these equals add the square of AD; therefore the squares of CB, BD, DA are equal to twice the rectangle CB, BD, and the squares of AD, DC: but the square of AB is equal (47. 1.) to the squares of BD, DA, because the angle BDA is a right angle, and the square B

D

C

of AC is equal to the squares of AD, DC: therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD, that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB, BB.

A

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 therefore the square of AB is equal (12. 2.) to the squares of AC, CB, and twice the rectangle BC, CD: to these equals add the square of BC, and the squares of AB, BC are equal to the square of AC, and twice the square of BC, B and twice the rectangle BC, CD: but because BD is divided into two parts in C; the rectangle DB, BC is equal (3. 2.) to the rectangle BC, CD and the square of BC: and the doubles of these are equal: therefore the squares of AB, BC are equal to the squares of AC, and twice the rectangle DB, BC: therefore the square of AC alone is less than the square of AB, BC by twice the rectangle DB, BC.

C

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 the square of AB, BC are equal (47. 1.) to the square of AC and twice the square of BC. Therefore, in every triangle, &c. Q. E. D.

D

Α

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

*See Note.

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 at 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 HE, EG are equal (47. 1.) to the square of GH; therefore the rectangle BE, EF together with the square of EG, is equal to the squares of HE, EG,: to 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 the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH; but BD is 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. Which was to be done.

THE

ELEMENTS OF EUCLID.

BOOK III.

DEFINITIONS.

I.

EQUAL circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumferences are equal.

'This is not a definition, but a theorem, the truth of which is evident; for, if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centre are equal.'

VII.

"The angle of a segment is that which is contained by the straight line and the circumference."

VIII.

An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment to the extremities of the straight line which is the base of the segment.

IX.

And an angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle.

X.

The sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them.

XI.

Similar segments of a circle, are

those in which the angles are

equal, or which contain equal angles.

PROP. I. PROB.

To find the centre of a given circle.*

Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisect (10. 1.) it in D; from the point D draw (11. 1.) DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the circle ABC.

For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are drawn from the centre Gt therefore the angle ADG is equal (8. 1.) to the angle GDB: but when a straight line standing upon another straight line makes the ad

* See Note.

+N. B. Whenever the expression "straight lines from the centre," or "drawn from the centre," occurs, it is to be understood that they are drawn to the cir cumference.