Putting s=(a+b+c) we have 28=a+b+c, and hence 28-2a=b+c-a, and therefore s-a= (b+c-a); similarly, 8 − b = 1 (a + c −b) and 8- - c = (a + b−c).

We shall then have from schol. 2, AB + CP = s, and CP = s-c; from schol. 3, AF or AG=s, and BF = BE = CP = CT=8-c. From schol. 4, if the angle C is right, PC + CT = diameter, or diameter = a+b-c; and therefore radius = (a + b −c) = s-c; also r's, in a right-angled triangle only. From schol. 5, the area of ABC= rs. From schol. 6, we have ABC=r' (s-a); and for the other two sides, area of ABC equals r" (s-b), and "" (sc).

S. Denoting the area of ABC by e, we have the following expressions for the radii of the several circles :

(4), go'' = (5), g

8-0

8 8-b

Hence dividing (3) by (4), (3) by (5), etc., we havo

(6);

that is, the reciprocal of the radius of the inscribed circle is equal to the sum of the reciprocals of the radii of the three ex-scribed circles.

If the perpendiculars from the angles on a, b, c be denoted respectively by p', p", p"", we have from (7),

In order to compare the perpendiculars with the radius of the inscribed circle, subtract the terms of each of the equations (7)

from those of the identical equation

=

; we thus obtain

PROPOSITIONS.

PROP. A.—THEOREM. (EUC. VI. D.)

The rectangle under the sides of a triangle is equal to the rectangle under the perpendicular to the base and the diameter of the circumscribing circle.

Let ACB be a triangle and CD a perpendicular on the base AB; then shall the rectangle contained by AC and CB be equal to that contained by A CD and the diameter EC of the circumscribing circle AEBC.

From CA cut off CO equal to CB, and CP equal to CD; join O, P and A, E, also E, B.

Then the angles ACE and ABE are equal, being in the same segment whose base is AE; and in the triangle DBC, D being right,

C and B are equal to a right angle,

E

D

B

(III. 12, cor. 2.)

but the angle EBC, in a semicircle, is a right angle. (III. 12, cor. 1.)

Take away the common angle ABC

and the remainder ABE is equal to BCD;

therefore ACE, that is, OCP is equal to BCD.

Now the sides OC and CP are equal to BC and CD,
and the contained angles have been proved equal;
therefore the angle CPO is equal to CDB,
and is therefore a right angle;

hence, also, OPE is right.

But the angle CAE is right;

therefore AOP and AEP together

are equal to a right angle,

and a circle might be described about the quadrilateral,

Hence the rectangle contained by AC and CO

so that CA and CE would then be secants.

is equal to that contained by EC and CP; that is, the rectangle under AC and CB

is equal to the rectangle under EC and CD.

(Ax. 3.) (Ax. 1.) (Const.)

(I. 4.)

(I. 10.)

(III. 12, cor. 1.)

(I. 23, cor. 1.) (III.12, cor.3.)

(III. 16.)

(Const.) Q. E. D.

Cor.-Hence calling the perpendicular p', and the radius of the circle R, we have a b=2 Rp'. (App. I., Schol. 7.)

Multiplying these equals by c, we have abc2c R p', but 2cp' = 4 (I. App., schol. 7); therefore R x 40=abc and

Rabc
40'

that is, the radius of the circumscribing circle is equal to the continued product of the sides divided by four times the

area.

PROP. B.-THEOREM.

In an equilateral triangle the same point is the centre of the inscribed and circumscribing circles, and the radius of the one is double that of the other.

K

E

Let ABC be an equilateral triangle. Then since the centre of the inscribed circle is at the point of meeting of the bisectors of the angles (IV. App., schol. 1), and since the straight line which bisects the angle of an equilateral triangle bisects the base at right angles, the centre of the inscribed circle must be at the point O on the perpendicular AF; but this point, by the same scholium, is the point of meeting of the perpendiculars from the middle of the sides and is the centre of the circumscribing circle. Also, it has been shewn (I. App. E, and Theor. 1, p. 73) that AF is trisected in the point O, so that each of the lines OF, OE, OK is one-third of the altitude of the triangle, and OA, OB, OC are each two-thirds of the same line.

B

Q. E. D.

EXERCISES ON BOOK IV.

1. If an equilateral triangle be described about a given circle, the straight lines joining the points of contact form an equilateral triangle, the side of which is half that of the circumscribed triangle.

2. If a straight line be drawn from an angle of a regular pentagon, to bisect the opposite side, it will pass through the centre of the circumscribing circle.

3. To inscribe a circle in a given sector of a circle.

4. In a given circle to inscribe three equal circles touching one another and the given circle.

5. The square on the side of a regular pentagon, inscribed in a circle, is equal to the squares on the sides of the regular hexagon and decagon inscribed in the same circle.

6. The square on the side of an equilateral triangle, inscribed in a circle, is equal to three times the square on the side of the regular hexagon, or radius of the circle.

7. To inscribe in an equilateral triangle three circles touching one another and each touching two of the sides.

8. Inscribe a square and a circle in a given quadrant.

9. The vertical angle of a triangle, inscribed in a circle, is greater or less than a right angle, by the angle contained by the base and the diameter drawn from the extremity of the base.

10. If from any point in the circumference of a given circle straight lines be drawn to the three angles of the inscribed equilateral triangle, the one drawn to the remote angle is equal to the sum of the other two.

11. The area of an inscribed regular hexagon is three-fourths that of the hexagon circumscribed about the same circle.

12. The least square which can be inscribed in a given square is that which is the half of the given square.

13. If through the vertex A of an equilateral triangle ABC, a perpendicular to the side AB be drawn to meet a perpendicular BD, drawn from the extremity B of the base, it shall be equal to the radius of the circle circumscribing the triangle.

14. If a circle be inscribed in, and one described about, a given right-angled triangle, the sum of their diameters is equal to the sum of the sides containing the right angle.

15. The straight line bisecting any angle of a triangle inscribed in a circle, cuts the circumference in a point which is equidistant from the centre of the circle, and from the extremities of the side opposite to the bisected angle.

16. The square inscribed in a circle is equal to half the square upon its diameter.

17. In a given circle to inscribe a rectangle equal to a given rectilineal figure, not exceeding half the square upon the diameter.

18. In a given circle to inscribe four equal circles touching one another and the given circle.