the distance between its extremity and the intersection of the opposite side, produced to meet it; and the angle formed by the intersection of the diagonals; to construct the trapezium.

With a diameter equal to the given longest side, describe a circle, and from its centre draw the radii

OC, OD making with each other an angle equal to twice. the complement of the given angle formed by the intersection of the diagonals. Join CD, and produce it; and with the centre O, and radius equal to the radius of the circle together with the given intercepted distance, describe a circle cutting it in E. Join EO, and produce it to B; AE is equal to the given intercepted distance. Join BC, AD; ABCD is the trapezium required.

Join AC, BD. Then the angle ACB in a semicircle being a right angle, FBC is the complement of CFB; but FBC is half of DOC, and .. BFC is equal to the given angle to be made by the diagonals.

(70.) The diagonals of a quadrilateral figure inscribed in a circle are to one another as the sums of the rectangles of the sides which meet their extremities.

Let ABCD be a quadrilateral figure inscribed in a circle; join AC, BD; AC is to BD, as the rectangles AB, AD and CB, CD together, to the rectangles AB, BC and AD, DC together.

H

F

Make the angle ABF equal to the angle DBC; to each of which add the angle FBD,.. the angle ABD is

equal to EBC; and ADB is equal to ECB, being in the same segment; .. the triangles ABD, EBC are equiangular, and

=

AB : BD :: BE : BC,

..the rectangle BD, BE is equal to the rectangle AB, BC. Join FC; and since the angle ABD = FBC, .. AD FC. Also since the angle EFC is equal to BDC in the same segment, and ECF equal to ABF, i. e. to DBC; .. the triangles ECF, BDC are equiangular, and BD DC :: CF: FE,

.. the rectangle FE, BD is equal to the rectangle CF, CD, i. e. to the rectangle AD, DC; whence the rectangles BE, BD and FE, BD or the rectangle BF, BD is equal to the rectangles AB, BC and AD, DC together. In the same manner if the angle BCG be taken equal to the angle ACD, it may be shewn that the rectangle CG, CA is equal to the rectangles AB, AD and CB, CD together. And since the angle BCG=ACD, the arc BG is equal to AD, i. e. to FC; to each of these add GF; .. the arc BAF is equal to GFC, and consequently the line BFGC; .. the rectangle AC, CG is equal to the rectangle AC, BF. And AC: BD as the rectangle AC, BF to the rectangle BD, BF, i. e. as the rectangles AB, AD and CB, CD together, to the rectangles AB, BC and AD, DC together.

(71.) The square described on the side of an equilateral and equiangular pentagon inscribed in a circle, is equal to the sum of the squares of the sides of a regular hexagon and decagon inscribed in the same circle.

Let ABC be an isosceles triangle having each of the

E

angles at the base double of the angle at A. With the centre A, and radius AB, describe a circle BCE. Draw CE bisecting the angle ACB. Join EB, EA; and draw EF perpendicular to AB. Then the angle EAB is double of ECB, and therefore is equal to CDB (Eucl. iv. 10.), and consequently is equal to the verti cally opposite angle ADE; whence AE-ED. Hence EB, ED and BC are equal to the sides of a regular pentagon, hexagon and decagon, respectively inscribed in the circle; and the squares of BC and DE are together equal to the square of BE.

For the angles at F being right angles, the squares of AF, FE are equal to the square of AE, i. e. to the square of ED or to the squares of EF, FD; whence AF is equal to FD. And since AD is bisected in F, and produced to D, the rectangle AB, BD together with the square of DF is equal to the square of BF; .. the rectangle AB, BD together with the squares of DF and FE, is equal to the squares of BF and FE; or the rectangle AB, BD together with the square of DE is equal to the square of BE. But (Eucl. iv. 10.) the rectangle AB, BD is equal to the square of AD, i. e. to the square of BC; .. the squares of BC, DE are together equal to the square of BE.

(72.) If the opposite sides of an irregular hexagon inscribed in a circle be produced till they meet; the three points of intersection will be in the same straight line.

Let ABCDEF be the hexagon inscribed in the circle; and let its opposite sides meet in G, H, I. Join

two opposite angles as A, D;

and about ADI describe a circle meeting GA, GD produced if necessary, in K, L. Join FC, KI, LI, LK. Then because AKID is a quadrilateral figure inscribed in a circle, the angle AKI is equal to ADE: and for the same reason ADE is equal to GFE, .. the angle AKI is equal to GFE, and KI is parallel to FH. In the same

manner it may be shewn that the angle AIL is equal to ADL and consequently to CB1, and LI parallel to HB. Again the angle KLD is equal to DAF, i. e. to FCD, and KL is parallel to FC.

.. GF : FH :: GK: KI;

whence G, H, I will be in a straight line.

SECT. VII.

(1.) THE vertical angle of any oblique-angled 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.

Let ABC be a triangle inscribed in a circle. From A draw the diameter AD; join BD; the angle ABC

is greater or less than a right angle, by the angle CAD.

For the angle ABD in a semicircle is a right angle; and ABC is equal to the

A

B

sum of ABD and DBC in one case; and is equal to their difference in the other; and in each case DBC= DAC in the same segment.

(2.) If from the vertex of an isosceles triangle a circle be described with a radius less than one of the equal sides, but greater than the perpendicular; the parts of the base cut off by it, will be equal.

From the vertex O of the isosceles triangle AOB, with a radius less than 40, but greater than the perpendicular from O on AB, let a

E

circle be described, cutting AB in C and D ; AC=BD. Join EF, OC, OD.

Then OE OB :: OF: OA,

:

and. EF is parallel to AB; and (ii. 1.) the arc FC equal to the arc DE, or the angle FOC=DOE; but AO, OC are equal to BO, OD each to each; .. AC= DB.

(3.) If a circle be inscribed in a right-angled triangle; the difference between the two sides containing the right angle and the hypothenuse, is equal to the diameter of the circle.

G G