The geometry, by T. S. Davies. Conic sections, by Stephen Fenwick

(32.111.) to the angle in the alternate segment BAC of the circle:

But the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D. Wherefore the segment BAC is cut off from the given circle ABC, containing an angle equal to the given angle D. Which was to be done.

PROPOSITION XXXV.

THEOR. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E; the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED.

If AC, BD pass each of them through the centre, so that E is the centre, it is evident that AE, EC, BE, ED being all (15 Def. 1) equal, the rectangle AE, EC is likewise equal to the rectangle BE, ED.

But let one of them, BD, pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E: then, if BD be bisected in F, F is the centre of the circle ABCD; join AF And because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, AE, EC are equal (3. III.) to one another:

And because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED, together with A the square of EF, is equal (5. 11.) to the square of FB; that is, to the square of FA :

But the squares of AE, EF are equal (47. 1.) to the square of FA; therefore the rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF:

Take away the common square of EF, and the remaining rectangle BE, ED is equal (3 Ax.) to the remaining square of AE; that is, to the rectangle AE, EC.

Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre in E, but not at right angles: then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (12. 1.) FG perpendicular to AC;

Therefore AG is equal (3. III.) to GC; wherefore the rectangle AE, EC, together with the square of EG, is equal (5. 11.) to the square of AG:

To each of these equals add the square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to the squares of AG, GF:

But the squares of EG, GF are equal (47. 1.) to the square of EF; and the squares of AG, GF are

equal to the square of AF; therefore the rectangle AE, EC, together with the square of EF, is equal to the square of AF; that is, to the square of FB:

But the square of FB is equal (5. 11.) to the rectangle BE, ED, together with the square of EF; therefore the rectangle AE, EC, together with the square of EF, is equal to the rectangle BE, ED, together with the square of EF:

Take away the common square of EF, and the remaining rectangle AE, EC is therefore equal to the remaining rectangle BE, ED.

Lastly, let neither of the straight lines AC, BD pass through the centre: take (1. III.) the centre F, and through E, H

the intersection of the straight lines AC, DB, draw

the diameter GEFH:

And because the rectangle AE, EC is equal, as has been shown, to the rectangle GE, EH; and, for the same reason, the rectangle BE, ED is equal to the same rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED. Wherefore, if two straight lines, etc. Q. E. D.

PROPOSITION XXXVI.

THEOR. If from any 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 of the line which touches it.

Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts the circle, and DB touches the same: the rectangle AD, DC is equal to the square of DB.

Either DCA passes through the centre, or it does not: first, let it pass through the centre E, and join EB; therefore the angle EBD is a right (18. III.) angle:

And because the straight line AC is bisected in E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equal (6. 11.) to the square of ED and CE is equal to EB; therefore the rectangle AD, DC, together with the square of EB, is equal to the square of ED :

But the square of ED is equal (47. 1.) to the squares of EB, BD, because EBD is a right angle; therefore

the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD:

Take away the common square of EB; therefore the remaining rectangle AD, DC is equal to the square of the tangent DB.

But if DCA does not pass through the centre of the circle ABC, take (1. III.) the centre E, and draw EF perpendicular (12. 1.) to AC, and join EB, EC, ED.

And because the straight line EF, which passes through the centre, cuts the straight line, AC, which does not pass through the centre, at

right angles, it shall likewise bisect (3.111.) it; therefore AF is equal to FC:

And because the straight line AC is bisected in F, and produced to D, the rectangle AD, DC, together with the square of FC, is equal (6. 11.) to the square of FD:

To each of these equals add the square of FE; therefore the rectangle AD, DC, together with the squares of CF, FE, is equal to the squares of DF, FE:

But the square of ED is equal (47. 1.) to the squares of DF, FE, because EFD is a right angle: and the square of EC is equal to the squares of CF, FE; therefore the rectangle AD, DC, together with the square of EC, is equal to the square of ED :

And CE is equal to EB; therefore the rectangle AD, DC, together with the square of EB, is equal to the square of ED :

But the squares of EB, BD are equal to the square (47. 1.) of ED, because EBD is a right angle; therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD:

Take away the common square of EB; therefore the remaining rectangle AD, DC is equal to the square of DB. Wherefore, if from any point, etc. Q. E. D.

COR. 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 D of them without the circle are equal to one another, viz., the rectangle BA, AE, to the rectangle CA, AF; for each of them is equal to the square of the straight line AD, which touches the circle.

PROPOSITION XXXVII.

THEOR. If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; 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 of the line which meets it, the line which meets shall touch the circle.

Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn of which DCA cuts the circle, and DB meets it: if the rectangle AD, DC be equal to the square of DB, DB touches the circle.

Draw (17. 111.) the straight line DE, touching the circle ABC; find (1. III.) its centre F, and join FE, FB, FD; then FED is a right (18. III.) angle:

And because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal (36. 11.) to the square of DE:

But the rectangle AD, DC is, by hypothesis, equal to the square of DB; therefore the square of DE is equal to the square of DB:

And the straight line DE equal to the straight line DB: and FE is

equal to FB; wherefore DE, EF are equal to DB, BF; and the base FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal (8. 1.) to the angle DBF:

But DEF is a right angle; therefore also DBF is a right angle and FB, if produced, is a diameter; and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches (16. 111.) the circle: therefore DB touches the circle ABC. Wherefore, if from a point, etc. Q. E. D.

EXERCISES ON BOOK III.

1. Let two parallel lines meet a circle: then,

(a.) If these both touch the circle, the points of contact are at the opposite extremities of a diameter;

(b.) If one cut and the other touch, the first cuts off equal arcs of the circle estimated from the point of contact of the second ; (c.) If both cut, they intercept equal arcs of the circle between them;

(d, e, f) State and prove the converses of these propositions. 2. If two lines be oblique to each other, and parallels to them be drawn to cut a circle: then,

(a.) The sum of the two opposite arcs, intercepted between these two lines, will be always the same, wherever the point of intersection may be, provided it be within the circle;

(b.) The difference will be the same wherever the point of intersection be taken without the circle;

(c, d.) State and prove the converses.

3. Two triangles ABC, A'B'C' are inscribed in the same circle, and the sides AB, AC are respectively parallel to A'B', A'C'; show that B'C is parallel to BC', and BB' equal to CC'.

4. Two chords of a circle cut one another and make equal angles with the diameter through the point of intersection: these chords will be equal.

5. If two chords AB, CD cut one another at right angles at E: then,

(a.) The sum of the arcs AC, BD are together equal to AD, BC; and each pair together equal to a semicircle;

(b.) The sum of the squares on AE, BE, CE, DE, is equal to the square on the diameter.

6. (a.) Find the least and greatest chord that can be drawn through a given point within a circle:

(b.) Through a given point within or without a circle draw a chord to cut off a quadrant :

(c.) Also through that point draw a chord of given length; and state the limitations of position to which that point must be subject.

7. In a given line to find a point such that tangents drawn from it to a given circle shall contain a given angle. Is this capable of one or more solutions?

8. If equal chords be drawn in a circle, they will all be tangents to another circle concentric with the former; and these chords will all be bisected at their points of contact.

Conversely, if chords be drawn in the outer of two concentric circles to touch the inner, these will all be equal, and be bisected at the points of contact.

9. Given the base and vertical angle of a triangle to construct it in the several cases where the third datum is:

(a.) The perpendicular;
(b.) The sum of the sides;

(c.) The difference of the sides;

(d.) One of the angles at the base;

(e.) The difference of the angles at the base;

(f.) A line drawn from the vertex to a given point in the base.

10. Given the angles at the base of a triangle and the radius of the circumscribing circle: to construct the triangle; and to point out which of the cases of the preceding exercise is virtually identical with this.

11. Except one pair of opposite angles together be equal to the other pair together in a quadrilateral, a circle cannot be described about the quadrilateral; but if that condition be fulfilled, it always

can.

12. No parallelogram but a rectangle can be inscribed in a circle. 13. If two circles touch each other, any straight line drawn through the point of contact will divide them into segments, of which the greater is similar to the greater and the less to the less; and the diameters through the points of intersection of the line with the circles will be parallel.

14. Through two given points to describe a circle which shall bisect the circumference of a given circle.

15. If all the interior or all the exterior angles of a quadrilateral be bisected, and the four bisecting lines in each case form a new quadrilateral then each of these quadrilaterals will be capable of circumscription by a circle.

16. If about each of the triangles into which a quadrilateral is divided by its diagonals, circles be described; their centres will be the angular points of a parallelogram.

What takes place when the given quadrilateral is inscriptible in a circle; and when it is a parallelogram?

17. Let the opposite sides of a quadrilateral be produced to meet, and circles be described about the four triangles of the figure which have these points for their vertices: then these circles all meet in one point, and their centres will be in one circumference.

18. A tower of given height is built on the side of a hill whose angle of inclination to the horizon is given; and a flagstaff of given length surmounts the tower: where must the spectator stand (the height of his eye above the ground also given) to see the flagstaff under a given angle, and under the greatest angle possible?

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