B

D

the centre E, and draw (I. 12.) EF perpendicular to AC, and join EB, EC, ED. Then, because the straight line EF, which passes through the centre, is perpendicular to the chord AC, AF is equal (III. 3.) to FC. And (II. 6.) because AC is bisected in F, and produced to D, the rectangle AD.DC and the square of FC are equal to the square of FD. To each of these equals add the square of FE; therefore the rectangle AD.DC, and the squares of CF, FE are equal to the squares of DF, FE: but (I. 47.) the square of ED is equal to the squares of DF, FE, because EFD is a right angle; and the square of EC, or (I. 46. cor. 2.) of EB, is equal to the squares of CF, FE: therefore the rectangle AD.DC and the square of EB are equal to the square of ED. But (I. 47.) the squares of EB, BD are equal to the square of ED, because EBD is a right angle; therefore the rectangle AD.DC, and the square of EB are 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, &c.*

E

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 of them without the circle, are equal to one another, viz., the rectangle BA.AE, to the rectangle CA.AF: for each rectangle is equal to the square of the tangent AD.

F

D

E

B

PROP. 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 the circle touches it.

If from a point without the circle ABC, two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it; and if the rectangle AD.DC be equal to the square of DB; DB touches the circle.

*The second case may be demonstrated more briefly thus: Join AE. Then the rectangle AD.DC is equal (II. 5. cor. 5.) to the difference of the squares of ED and EC, or of ED and EB, or (I. 47.) to the square of DB.

But the

E

Draw (III. 17.) the straight line DE touching the circle ABC; find (III. 1.) the centre F; and join FE, FB, FD. Then (III. 18.) FED is a right angle: and (III. 36.) because DE touches the circle ABC, and DCA cuts it, the rectangle AD.DC is equal to the square of DE. 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 (I. 46. cor. 3.) to the straight line DB. But FE is equal to FB, and the base FD is common to the two triangles DEF, DBF; therefore (I. 8.) the angle DEF is equal to DBF; but DEF is a right angle; therefore also DBF is a right angle: and FB is part of a diameter, and the straight line which is drawn at right angles to a diameter, from its extremity, touches (III. 16. cor.) the circle: therefore DB touches the circle ABC. Wherefore, if from a point, &c.

A

BOOK IV.*

DEFINITIONS.

1. WHEN the sides of one rectilineal figure pass through the angular points of another, the figures. not coinciding with one another, the interior figure is said to be inscribed in the exterior, and the exterior to be circumscribed, or described, about the interior one.

2. When all the angular points of a rectilineal figure are upon the circumference of a circle, the rectilineal figure is said to be inscribed in the circle, and the circle to be circumscribed, or described, about the rectilineal figure.

3. When each side of a rectilineal figure touches a circle, the rectilineal figure is said to be circumscribed, or described, about the circle, and the circle to be inscribed in the rectilineal figure.

PROP. I. PROB.

In a given circle to place a chord equal to a given straight line, not greater than the diameter.

Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle.

C

A

Draw BC a diameter of the circle ABC; then, if BC be equal to D, the thing required is done; for in the circle ABC a chord BC is placed equal to D. But if it be not, BC is greater than D; make CE equal (I. 3.) to D, and from the centre C, at the distance CE, describe (I. post. 3.) the circle AEF, and join CA. Then, because C is the centre of the circle AEF, CA is equal to CE; but D is equal to CE; therefore D is equal to CA: wherefore in the circle

D

F

B

E

This book consists of problems, chiefly respecting the rectilineal figures described in circles or about them.

ABC, a chord is placed equal to the given straight line D; which was to be done.

PROP. II. PROB.

In a given circle to inscribe a triangle equiangular to a given triangle.

Let ABC be a given circle, and DEF a given triangle; it is required to inscribe in ABC a triangle equiangular to DEF. Draw the straight line GAH (III. 17.) touching the circle in any point A, and make (I. 23.) the angle HAC equal to E, and GAB equal to F, and join BC: ABC is the triangle required.

E

D

B

H

For, since GAH touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal (III. 32.) to the angle B in the alternate segment; but HAC is equal (const.) to E; therefore also B is equal to E. For the same reason, the angle C is equal to F; therefore the remaining angle BAC is equal (I. 32. cor. 5.) to the remaining angle D: wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed (IV. def. 2.) in the circle ABC: which was to be done.

PROP. III. PROB.

ABOUT a given circle to describe a triangle equiangular to a given triangle.

Let ABC be a given circle, and DEF a given triangle; it is required to describe a triangle about ABC, equiangular to DEF. Produce EF both ways to G, H, and find (III. 1.) K the centre of the circle; from it draw any radius KB; and make (I. 23.) the angle BKA equal to DEG, and BKC equal to DFH; through the points A, B, C, draw (III. 17.) the tangents LAM, MBN, NCL LMN is the required triangle. *

A

D

GE

FH

K

M B

N

For, since LM, MN, NL touch the circle ABC in the points A, B, C, to which KA, KB, KC are drawn from the centre, the angles at the points A, B, C are (III. 18.) right angles and because the four angles of the quadrilateral figure AMBK are equal (I. 32. cor. 1.) to four right angles and that two of them KAM, KBM are right

:

That ML and NL must meet, might be shown by joining AC; for the angles CAL, ACL being less than the right angles at A and C, AL and CL (1. ax. 12.) must meet. In a similar manner it might be proved that LM and NM, and also LN and MN, must meet.

GE

D

angles, the other two AKB, AMB are equal to two right angles. But (I. 13.) the angles DEG, DEF are likewise equal to two right angles; therefore the angles AKB, AMB are equal to the angles DEG, DEF of which (const.) AKB is equal to DEG: wherefore the remaining angles AMB, DEF are equal. In like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal (I. 32. cor. 5.) to the remaining angle EDF: wherefore the triangle LMN is equiangular to DEF: and it is described about the circle ABC: which was to be done.

PROP. IV. PROB.

M B

To inscribe a circle in a given triangle.

Let the given triangle be ABC; it is required to inscribe a circle in it.

E

D

G

Bisect (I. 9.) the angles ABC, BCA by the straight lines BD, CD, meeting one another in the point D; and from D draw (I. 12.) DE, DF, DG, perpendiculars to AB, BC, CA. Then, because the angle EBD is equal (const.) to FBD, and that the right angles BED, BFD are equal, the two triangles EBD, FBD have two angles of the one equal to two angles of the other, and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore (I. 26.) DE is equal to DF. In the same manner it would be shown, that, in the triangles DGC, B F DFC, DG is equal to DF: therefore the three straight lines DE, DF, DG are equal to one another, and the circle described from the centre D, at the distance of any of them, will pass through the extremities of the other two; and (III. 16. cor.) it will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right angles. Therefore the straight lines AB, BC, CA touch the circle, and the circle EFG is inscribed (IV. def. 3.) in the triangle ABC; which was to be done.

Schol. This proposition is a case of the general problem, to describe a circle touching three given straight lines which do not pass through the same point, and which are not all parallel to one another.

If two of the lines be parallel, there may evidently be two equal circles, one on each side of the line falling on the parallels, each of which will touch the three given lines and their centres will be the intersections of the lines bisecting the angles made by the parallels with the third line.

If the lines form a triangle by their intersections, there will be