Then, because each of the angles BAC and BAG is a

to BA in the other, and the side BC in the one is equal to BD in the other, and the included angles are equal, as just proved.

Now, the area of the parallelogram BL is double that of the triangle DBA, because they have the same base BD, and the same altitude DL (389). And the area of the square BG is double that of the triangle FBC, because these also have the same base BF, and the same altitude FG. But doubles of equals are equal (7). Therefore, the parallelogram BL and the square BG are equivalent.

In the same manner, by joining AE and BK, it is demonstrated that the parallelogram CL and the square CH are equivalent. Therefore, the whole square BE, described on the hypotenuse, is equivalent to the two squares BG and CH, described on the legs of the right angled triangle.

409. Corollary.-The square described on one leg is equivalent to the difference of the squares on the hypotenuse and the other leg.

410. If from the extremities of one line perpendiculars be let fall upon another, then the part of the second line between the perpendiculars is called the projection of the first line on the second. If one end of the first line is in the second, then only one perpendicular is necessary.

411. Theorem.-The square described on the side opposite to an acute angle of a triangle, is equivalent to the sum of the squares described on the other two sides, diminished by twice the rectangle contained by one of these sides and the projection of the other on that side.

Let A be the acute angle, and from B let a perpendicular fall upon AC, produced if necessary. Then,

AD is the projection of AB upon AC. And it is to be proved that the square on BC is equivalent to the sum of the squares on AB and on AC, diminished by twice the rectangle contained by AC and AD.

But the square on BC is equivalent to BD+CD (408).

Therefore, it is also equivalent to

412. Theorem.—The square described on the side opposite an obtuse angle of a triangle, is equivalent to the sum of the squares described on the other two sides, increased by twice the rectangle of one of those sides and the projection of the other on that side.

B

In the triangle ABC, the square on BC which is opposite the obtuse angle at A, is equivalent to the sum of the squares on AB and on AC, and twice the rectangle contained by CA and

By addition, BD+CD=AB+ AC+2AC X AD.

413. Corollary.-If the square described on one side of a triangle is equivalent to the sum of the squares described on the other two sides, then the opposite angle is a right angle. For the last two theorems show that it can be neither acute nor obtuse.

EXERCISES.

414.-1. When a quadrilateral has its opposite angles supplementary, a circle can be circumscribed about it.

2. From a given isosceles triangle, to cut off a trapezoid which

shall have the same base as the triangle, and the remaining three sides equal to each other.

3. The lines which bisect the angles of a parallelogram, form a rectangle whose diagonals are parallel to the sides of the parallelogram.

4. In any parallelogram, the distance of one vertex from a straight line passing through the opposite vertex, is equal to the sum or difference of the distances of the line from the other two vertices, according as the line is without or within the parallelogram.

5. When one diagonal of a quadrilateral divides the figure into equal triangles, is the figure necessarily a parallelogram?

6. Demonstrate the theorem, Article 329, by Articles 113 and 387.

7. What is the area of a lot, which has the shape of a right angled triangle, the longest side being 100 yards, and one of the other sides 36 yards.

8. Can every triangle be divided into two equal parts? Into three? Into nine?

9. Two parallelograms having the same base and altitude are equivalent.

To be demonstrated without using Articles 379 or 383.

10. A triangle is divided into two equivalent parts, by a line from the vertex to the middle of the base.

To be demonstrated without the aid of the principles of this chapter.

11. To divide a triangle into two equivalent parts, by a line drawn from a given point in one of the sides.

12. Of all equivalent parallelograms having equal bases, what one has the minimum perimeter?

13. Find the locus of the points such that the sum of the squares of the distances of each from two given points, shall be equivalent to the square of the line joining the given points.

CHAPTER VII.

POLYGONS.

415. Hitherto the student's attention has been given to polygons of three and of four sides only. He has seen how the theories of similarity and of linear ratio have grown out of the consideration of triangles; and how the study of quadrilaterals gives us the principles for the measure of surfaces, and the theory of equivalent figures.

In the present chapter, some principles of polygons of any number of sides will be established.

A Pentagon is a polygon of five sides; a Hexagon has six sides; an Octagon, eight; a Decagon, ten; a Dodecagon, twelve; and a Pentedecagon, fifteen.

The following propositions on diagonals, and on the sum of the angles, are more general statements of those in Articles 340 to 346.

DIAGONALS.

416. Theorem.-The number of diagonals from any vertex of a polygon, is three less than the number of sides.

For, from each vertex a diagonal may extend to every other vertex except itself, and the one adjacent on each side. Thus, the number is three less than the number of vertices, or of sides.

417. Corollary. The diagonals from one vertex di