tending 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 perpendicular to BC produced. Then the square of AB is greater than the squares of AC, CB, by twice the rectangle BC, CD.

C D

Because BD is divided in C, the square of BD is equal to the squares of BC, CD, and twice the rectangle BC, CD; to each add the square of DA; then 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 to the squares of BD, DA, since the angle at D is a right angle; and the square of CA is equal 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.

XIII.—In every triangle, the square of the side subtending either 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 acute angle and the perpendicular let fall upon it from the opposite 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 AD from the opposite angle. Then the square of AC opposite to the angle B, is less than the squares of CB, BA, by twice the rectangle CB. DB.

B .D

First, let AD fall within the triangle ABC. Then because the straight line CB is divided in D, the squares of CB, BD are equal to twice the rectangle contained by CB, BD, and the square of DC; to each 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 to the squares of BD, DA, because the angle BDA is a right angle; and the square of AC is equal to the squares of AD, DC; therefore the squares CB, BA are equal to the square of AC, and twice the rectangle CB, DB; that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB, BD. Next, let AD fall outside the triangle ABC. Then, because D is a right angle, the angle ACB is greater than a right angle; and the square of AB is equal to the squares

D

of AC, CB, and twice the rectangle B C BC, CD; to cach add the square of BC; therefore the squares of AB, BC are equal to the square of AC, twice the square of BC, and twice the rectangle BC, CD; but because BD is divided in C, the rectangle DB, BC is equal to the rectangle BC, CD, and the square of BC; and twice the rectangle DB, BC is equal to the rectangle BC, CD, and twice the square of BC: therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: therefore the square of AC is less than the squares of AB, BC, by twice the rectangle DB, BC. Lastly, let AC be perpendicular to BC. Then BC is the straight line between the perpendicular and the acute angle at B; and the squares AB, BC, are equal to the square of AC, and twice the square

BC.

of

B

XIV. To describe a square equal to a given rectilineal figure.

F

B

G E

D

Let A be the given rectilineal figure. Describe the rectangular parallelogram BCDE equal to A. Then, if the sides of it are equal to one another, it is a square. But if not, produce ВE to F, and make EF equal to ED, bisect BF in G; from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to meet the circumference in H. The square described upon EH shall be equal to the given rectilineal figure A. Join GH. Because BF is bisected in G, and divided into two unequal parts in E; the rectangle BE, EF, together with the square of EG, are equal 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 to the square of GH; therefore the rectangle BE, EF, together with the square of EG, are equal to the squares of EG, EH; take away the common square of EG; therefore the rectangle BE, EF is equal to the square of HE. 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 square of EH is equal to the rectilincal figure A.

ORIGINAL PROPOSITIONS RIDERS OR DEDUCTIONS.

As an exercise on the preceding propositions, and to develop the reasoning powers of the student, original propositions, riders, or deductions based on Euclid, are generally given to the student. These are to be solved by the pupil, who is to take nothing for granted but what is, or can be, proved from Euclid's Elements.

Such propositions are, as in Euclid, either Problems or Theorems. The best general rule that can be given for the solution of these is to consider the construction required drawn, or the proposition which has to be proved, true; and then to see what consequences will follow from this supposition.

If such consequences are seen to be conditions that can be proved from Euclid, the student then proceeds to reason in the reverse way from these conditions to the construction or proposition required.

This will be made clearer by reference to a few examples.

:

A. To bisect a given square. Let A B C D be the given square it is required to bisect it. Let us suppose this is done, and that the bisecting line passes diagonally through the square from A to C. If this is the case, the consequence will be that the triangle A B C is equal to the triangle C D A. But this follows from Euclid. Therefore, working reversely, from a draw the diagonal A 0; this divides the square, which is a parallelogram, into two equal parts, which was required to be done.

Another way. Let the section pass through the point of bisection of opposite sides. This divides the square into two parallelograms. But these are upon equal bases, and between the same parallels, therefore they are equal, and the original figure has been bisected.

In Euclid I. 1, if D, the lower point of intersection of the circles, be joined with A and B, the extremities

of the base of the equilateral triangle A B C prove that the figure A B C D is a parallelogram. Suppose the proposition be true, then a c is parallel to B D, and A D to CB, and the angle C A B is equal to the alternate angle B A D.

It can be proved, as in Euclid I. 1, that A B D is an equilateral triangle; therefore each of the angles CA B and A B D is one-third of two right angles, and these angles are therefore equal, and therefore the sides a c and B D are parallel: in the same way A D and C B are proved parallel to each other, and the figure A B C D is a parallelogram.