Corollary 1.-If from a point without a circle, two secants be drawn to the opposite circumference, the rectangles of the whole lines and the parts without the circle, will be equal.

the

=

Draw AEF; then AC. AB AF. AE, for each is equal to square of AD.

Corollary 2.-Two tangents drawn from the same point are equal.

Corollary 3.-If AB. AC-AE. AF, then the four points B, E, F, C are on the circumference of the same circle.

For, if not, let B, E, F, and H be on the same circle. Let the student prove an impossibility.

Proposition 28.

Theorem.—If an angle of a triangle be bisected by a straight line which also cuts the base, the rectangle contained by the sides of the triangle, is equal to the rectangle contained by the segments of the base, together with the square of the line bisecting the angle.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; then

BA.AC-BD.DC+AD2.

Describe (III. 29) the circle ABC about the triangle ABC, and produce AD to the circumference in E, and join EC.

B

E

Then, because the angle BAD is equal to the angle EAC, and the angle ABD to the angle AEC, they being in the same segment, the triangles ABD, AEC are equiangular.

Theorem.-If from any angle of a triangle a perpendicular be drawn to the base, the rectangle contained by the sides of the triangle, is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

Let ABC be a triangle, and AD the perpendicular; then

=

BA.AC the rectangle DA and a diameter of the circumscribed circle.

Describe the circle ABC about the triangle ABC; draw the diameter AE, and join EC.

B

Because the right angle BDA is equal to the angle ECA in a semicircle, and the angle ABD to the angle AEC in the same segment, the triangles ABD, AEC are equiangular.

Theorem.-The rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides.

Let ABCD be any quadrilateral inscribed in a circle, and let AC, DB be drawn; then

Again, because the angle ABE is equal to the angle DBC, and the angle BAE to the angle BDC in the same segment, the triangle ABE is equiangular to the triangle DBC.

.. (V. 4),

... (V. 14),

BA: AE :: BD : DC; .

BA.DC=BD. AE.

But we have shown BC. DA = BD. EC;

BA.DC+BC.DA-BD.AE+BD. EC.

But (II. 1) BD. AE+BD. EC=BD. AC;

BA.DC+BC. DA=BD.AC.

I

EXERCISES.

1. To draw a line from the vertex of a triangle bisecting the triangle (V. 1, Cor. 2).

2. Two squares are to each other as the squares of their diagonals.

3. If a line be drawn parallel to the base of a triangle, all lines from the vertex to the base will be cut proportionally; and the segments of the base and of the parallel will be proportional to each other.

4. The perimeters of similar polygons are proportional to their homologous sides (IV. 16).

5. A line drawn parallel to the parallel sides of a trapezoid will cut the other sides proportionally.

6. If two triangles be on equal bases which are in the same straight line, and between the same parallels, any straight line parallel to their bases will cut off equal triangles and trapezoids.

7. Let A, B and C be three points in a straight line; and through C let any straight lines pass; the perpendiculars on these lines from A and B bear a constant ratio to each other.

8. If one of the parallel sides of a trapezoid be double the other, its diagonals trisect each other.

9. Trisect a given line by use of (V. 3).

10. In the figure in (II. 11) show that four other lines besides the given one are divided in extreme and mean

ratio.

is

11. The area of an equilateral triangle whose side is 1

V/3

4

12. If the arc AB be bisected in C, and D be any point of the circumference, then AD+DB: DC:: AB: AC (V. 30).

13. A circle is described about an equilateral triangle; from any point of the circumference straight lines are drawn to the

D

B

angles of the triangle; then one of these lines is equal to the sum of the other two.

14. On a given line to construct a rectangle which shall be equal to a given rectangle (V. 14).

15. If from a point without the circle two lines be drawn, one cutting the circle and the other meeting, and the rectangle contained by the whole cutting-line and the part without the circle be equal to the square of the line meeting it, the latter line touches the circle (V. 27).

16. The parallelograms EH and GF of (I. 40) are similar to the whole parallelogram BD.

17. The base of a triangle is 6; of the triangle are cut off toward the base by a line parallel to it; what is the length of the cutting-line (V. 19)?

18. The area of a triangle is equal to its perimeter multiplied by one half the radius of the inscribed circle.

For other practical problems in this book see Mensuration at the end of Book VI.