PROPOSITION 36.-THEOREM.

If from a point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square on the line which touches it.

(References-Prop. I. 12, 47; II. 6; III. 1, 3, 18.)

Hypothesis. Let D be any point without the circle ABC, and let DCA, DB, be two straight lines drawn from it, of which DCA cuts the circle, and DB touches it.

Sequence. The rectangle AD DC, shall be equal to the square on DB.

Case I-Let DCA pass through the centre E, and join EB.

Demonstration.-1. Then EBD is a right angle. (III. 18.)

2. And because the straight line AC is bisected in E, and produced to D, the rectangle AD DC, together with the square on EC, is equal to the square on ED. (II. 6.)

3. But EC is equal to EB;

4. Therefore the rectangle AD·DC, together with the square on EB, is equal to the square on ED.

D

A

5. But the square on ED is equal to the squares on EB, BD, because EBD is a right angle; (I. 47.)

6. Therefore the rectangle AD DC, together with the square on EB, is equal to the squares on EB, BD.

7. Take away the common square on EB;

8. Then the remaining rectangle AD DC, is equal to the square on DB. (ax. 3.)

Case II.-Let DCA not pass through

the centre of the circle ABC.

Construction. Take the centre E

(III. 1), and draw EF perpendicular to
AC (I. 12), and join EB, EC, ED.
1. Because the

Demonstration.
straight line EF, which passes through
the centre, cuts the straight line AC,
which does not pass through the centre, at
right angles, it also bisects it; (III. 3.)
2. Therefore AF is equal to FC.
3. And because the straight line AC is

B

F

bisected in F, and produced to D, the rectangle AD DC, together with the square on FC, is equal to the square on FD. (II. 6.) 4. To each of these equals add the square on FE;

5. Therefore the rectangle AD DC, together with the squares on CF, FE, is equal to the squares on DF, FE. (ax. 2.) 6. But the squares on CF, FE, are equal to the square on CE, because CFE is a right angle; (I. 47.)

7. And the squares on DF, FE, are equal to the square on DE;

8. Therefore the rectangle AD DC, together with the square on CE, is equal to the square on DE.

9. But CE is equal to BE;

10. Therefore the rectangle AD DC, together with the square on BE, is equal to the square on DE.

11. But the square on DE is equal to the squares on DB, BE, because EBĎ is a right angle. (I. 47.)

12. Therefore the rectangle AD DC, together with the square on BE, is equal to the squares on DB, BE.

13. Take away the common square on BE;

14. Then the remaining rectangle AD DC is equal to the square on DB. (ax. 3.)

Conclusion. Therefore, if from any point, &c. Q. E. D.

Corollary. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another; namely, the rectangle BA AE is equal to the rectangle CA AF; for each of them is equal to the square on the straight line AD, which touches the circle.

D

PROPOSITION 37.--THEOREM.

If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on the line which meets the circle, the line which meets the circle shall touch it.

(References-Prop. I. 8; III. 1, 16 cor., 17, 18, 36.)

Hypothesis.-Let any point D be taken without the circle ABC, and from it let two straight lines, DCA, DB, be drawn,

of which DCA cuts the circle, and DB meets it; and let the rectangle AD DC, be equal to the square on DB.

Sequence. Then DB shall touch the circle.

Construction.-1. Draw the straight line DE, touching the

circle ABC; (III. 17.)

2. Find F the centre (III. 1), and join FB, FD, FE.

Demonstration.-1. Then FED is a right angle. (III. 18.)

2. And because DE touches the circle ABC, and DCA cuts it, the rectangle AD DC is equal to the square on DE. (III. 36.)

3. But the rectangle AD DC is equal to the square on DB; (hyp.)

4. Therefore the square on DE is equal

to the square on DB; (ax. 1.)

B

5. Therefore the straight line DE is equal to the straight line DB.

6. And EF is equal to BF; (I. def. 15.)

7. Therefore the two sides DE, EF, are equal to the two sides DB, BF, each to each;

8. And the base DF is common to the two triangles DEF, DBF;

9. Therefore the angle DEF is equal to the angle DBF. (I. 8.) 10. But DEF is a right angle; (const.)

11. Therefore also DBF is a right angle. (ax. 1.)

12. And BF, if produced, is a diameter; and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches the circle; (III. 16, cor.)

13. Therefore DB touches the circle ABC.

Conclusion. Therefore, if from a point, &c. Q. E.D.

EXERCISES ON BOOK III.

Prop. 1 to 15.

1. Describe a circle passing through three given points. When will this be impossible.

2. With a given radius, describe a circle passing through a given point, and touching a given straight line.

3. If two circles cut each other, any two parallel straight lines drawn through the points of section to cut the circumferences, are equal.

4. Through either of the points of intersection of two circles, draw the greatest straight line, terminated both ways by the circumferences.

5. If a tangent be drawn parallel to the diameter of any circle, the square of the straight line drawn from the point of contact to any point in the diameter is equal to the sum of the squares of the segments of the diameter.

6. If in two equal circles which cut each other, two parallel diameters be drawn, the line joining the extremities of these diameters shall bisect the line joining the centres.

7. The difference of the squares of any two chords in a circle is equal to four times the difference of the squares of the perpendiculars on them from the centre.

8. Through one of the points of intersection of two circles, draw a straight line to the circumferences, and bisected in that point.

9. Of straight lines joining points in the circumferences of two concentric circles, the least is that which, when produced, passes through the centre.

10. With a given radius, to describe a circle, touching a given circle, and having its centre in a given straight line.

11. If in any two circles which touch each other, parallel diameters be drawn, an extremity of each diameter and the point of contact shall lie in the same straight line.

12. Through a given point within a given circle, draw the shortest chord.

Prop. 16 to 30.

13. With a given radius, describe a circle having its centre in a given straight line, and touching another given straight line.

14. Describe a circle which shall touch a given straight line in a given point, and also touch a given circle.

15. Given one leg of a right-angled triangle, and the radius of the inscribed circle, to describe the triangle.

16. AB is any diameter of the circle ABC, whose centre is O, and DCE is a tangent at C, terminated in D and E by the tangents AD and BE at the extremities of the diameter; show that the angles AOD and BOE are respectively equal to COD, and COE.

17. Any parallelogram described about a circle is equilateral. 18. With a given radius describe a circle, touching a given circle, and a given straight line.

19. Given a right-angled triangle, describe a circle passing through the right angle, touching the hypothenuse, and having its centre in one of the legs.

20. Divide a given circle into two segments, so that the angle contained in the one may be three times the angle contained in the other.

21. The straight lines which join the extremities of parallel chords in a circle are equal.

Prop. 31 to 37.

22. If one circle touch another internally, the radius of the one being half that of the other, any line drawn from the point of contact to the outer circumference is bisected by the inner.

23. If two circles touch each other, any line drawn through the point of contact to cut the circumferences, cuts off similar segments.

24. If from either point of intersection of two circles, two diameters be drawn, the straight line joining their extremities passes through the point of contact.

25. Find a point within a triangle at which the three sides shall subtend equal angles.

26. Draw a tangent to a given circle, so that the part of it between two given tangents may be equal to a given straight line.

27. Given the base of a triangle, the vertical angle, and the altitude, to construct the triangle.

28. Given the base of a triangle, the vertical angle, and the sum or difference of the sides, to construct the triangle.

29. If two circles cut each other, the line joining the points of intersection being produced, bisects their common tangent. 30. Given the perimeter of a triangle, the vertical angle, and the altitude, to construct the triangle.

31. If from the extremities of the base of a triangle, perpendiculars be drawn to the sides, the rectangle contained by the one side and its segment remote from the base, is equal to the rectangle contained by the other side and its segment similarly

situated.

32. Draw a common tangent to two given circles.