are produced to meet AD and AB produced in E and F. The angles ABC and ADC are together equal to AFC, AEB, and twice the angle BAC.

145. If the hypotenuse AB of a right-angled triangle ABC be bisected in D, and EDF drawn perpendicular to AB, and DE, DF cut off each equal to DA, and CE, CF joined, prove that the last two lines will bisect the angle at C and its supplement respectively.

146. ABCD is a quadrilateral figure inscribed in a circle. Through its angular points tangents are drawn so as to form another quadrilateral figure FBLCHDEA circumscribed about the circle. Find the relation which exists between the angles of the exterior and the angles of the interior figure.

147. The angle contained by the tangents drawn at the extremities of any chord in a circle is equal to the difference of the angles in segments made by the chord: and also equal to twice the angle contained by the same chord and a diameter drawn from either of its extremities.

148. If ABCD be a quadrilateral figure, and the lines AB, AC, AD be equal, shew that the angle BAĎ is double of CBD and CDB together.

149. If the sides of a quadrilateral figure circumscribing a circle, touch the circle at the angular points of an inscribed quadrilateral figure; all the diagonals will intersect in the same point.

150. In a quadrilateral figure ABCD is inscribed a second quadrilateral by joining the middle points of its adjacent sides; a third is similarly inscribed in the second, and so on. Shew that each

of the series of quadrilaterals will be capable of being inscribed in a circle if the first three are so. Shew also that two at least of the opposite sides of ABCD must be equal, and that the two squares upon these sides are together equal to the sum of the squares upon the other two.

XIV.

151. If from any point in the diameter of a semicircle, there be drawn two straight lines to the circumference, one to the bisection of the circumference, the other at right angles to the diameter, the squares upon these two lines are together double of the square upon the semi-diameter.

152. If from any point in the diameter of a circle, straight lines be drawn to the extremities of a parallel chord, the squares on these lines are together equal to the squares on the segments into which the diameter is divided.

153. From a given point without a circle, at a distance from the circumference of the circle not greater than its diameter, draw a straight line to the concave circumference which shall be bisected by the convex circumference.

154. If any two chords be drawn in a circle perpendicular to each other, the sum of their squares is equal to twice the square of the diameter diminished by four times the square of the line joining the center with their point of intersection.

155. Two points are taken in the diameter of a circle at any equal distances from the center; through one of these draw any chord, and join its extremities and the other point. The triangle so formed has the sum of the squares of its sides invariable.

156. If chords drawn from any fixed point in the circumference of a circle, be cut by another chord which is parallel to the tangent at that point, the rectangle contained by each chord, and the part of it intercepted between the given point and the given chord, is constant. 157. If AB be a chord of a circle inclined by half a right angle to the tangent at A, and AC, AD be any two chords equally inclined to AB, ACAD2 = 2. AB2.

158. A chord POQ cuts the diameter of a circle in Q, in an angle equal to half a right angle; PO + OQ2 = 2 (rad.)2.

159. Let ACDB be a semicircle whose diameter is AB; and AD, BC any two chords intersecting in P; prove that

ABDA. AP + CB.BP.

160. If ABDC be any parallelogram, and if a circle be described passing through the point A, and cutting the sides AB, AC, and the diagonal AD, in the points F, G, H respectively, shew that

AB.AF+AC.AG=AD.AH.

161. Produce a given straight line, so that the rectangle under the given line, and the whole line produced, may equal the square of the part produced.

162. If A be a point within a circle, BC the diameter, and through A, AD be drawn perpendicular to the diameter, and BAE meeting the circumference in E, then BA.BE=BC.BD.

163. The diameter ACD of a circle, whose center is C, is produced to P, determine a point F in the line AP such that the rectangle PF.PC may be equal to the rectangle PD. PA.

164. To produce a given straight line, so that the rectangle contained by the whole line thus produced, and the part of it produced, shall be equal to a given square.

165. Two straight lines stand at right angles to each other, one of which passes through the center of a given circle, and from any point in the other, tangents are drawn to the circle. Prove that the chord joining the points of contact cuts the first line in the same point, whatever be the point in the second from which the tangents are drawn.

166. A, B, C, D, are four points in order in a straight line, find a point E between B and C, such that AE. EB = EĎ. EC, by a geometrical construction.

167. If any two circles touch each other in the point O, and lines be drawn through O at right angles to each other, the one line cutting the circles in P, P, the other in Q, Q'; and if the line joining the centers of the circles cut them in A, A'; then P'P2 + Q'Q2 = A'A3.

DEFINITIONS.

I.

A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each.

II.

In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each.

III.

A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle.

IV.

A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle.

V.

In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure.

VI.

A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described.

о

VII.

A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.

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

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

It is required to place in the circle ABC a straight line equal to D.

Draw BC the diameter of the circle ABC. Then, if BC is equal to D, the thing required is done; for in the circle ABC a straight line BC is placed equal to D. But, if it is not, BC is greater than D; (hyp.) make CE equal to D, (1. 3.)

and from the center C, at the distance CE, describe the circle AEF, and join CA.

Then CA shall be equal to D.

Because C is the center of the circle AEF,
therefore CA is equal to CE: (I. def. 15.)
but CE is equal to D; (constr.)

therefore D is equal to CA. (ax. 1.)

Wherefore in the circle ABC, a straight line CA is placed equal to the given straight line D, which is not greater than the diameter of the circle. Q.E.F.

PROPOSITION II. PROBLEM.

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

Let ABC be the given circle, and DEF the given triangle. It is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.

Draw the straight line GAH touching the circle in the point A, (III. 17.) and at the point A, in the straight line AH,

make the angle HAC equal to the angle DEF; (1. 23.)
and at the point A, in the straight line AG,
make the angle GAB equal to the angle DFE;
and join BC: then ABC shall be the triangle required.
Because HAG touches the circle ABC,

and AC is drawn from the point of contact, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle: (III. 32.)

but HAC is equal to the angle DEF; (constr.)

therefore also the angle ABC is equal to DEF: (ax. 1.)

for the same reason, the angle ACB is equal to the angle DFE: therefore the remaining angle BAC is equal to the remaining angle EDF: (1. 32. and ax. 1.)

wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC.

PROPOSITION III. PROBLEM.

Q. E. F.

About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle. It is required to describe a triangle about the circle ABC equiangular to the triangle DEF.

Produce EF both ways to the points G, H;
find the center K of the circle ABC, (III. 1.)
and from it draw any straight line KB;

at the point K in the straight line KB,
make the angle BKA equal to the angle DEG, (1. 23.)
and the angle BKC equal to the angle DFH;

and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC. (III. 17.)

Then LMN shall be the triangle required.

Because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the center are drawn KA, KB, KC,

therefore the angles at the points A, B, C are right angles: (III. 18.) and because the four angles of the quadrilateral figure AMBK are equal to four right angles,

for it can be divided into two triangles;

and that two of them KAM, KBM are right angles, therefore the other two AKB, AMB are equal to two right angles : (ax. 3.)

but the angles DEG, DEF are likewise equal to two right angles;

(I. 13.)