parallel to AD, it is perpendicular to AB; .. the angle GKO is a right angle, and equal to AFO; and the vertical angles at O are equal, and GO=OA; .. the triangles GKO, OFA are equal, and OF=OK. But since OB=BE, .. (Eucl. vi. 2.) OK=KI; and .. OF = KI, and OG=KH; .. FG= IH.

(64.) If two sides of a trapezium inscribed in a circle be produced, and from the same point in one side produced a line be drawn parallel to the other, intersecting the adjacent side of the trapezium, and a second line to the extremity of that other intersecting the circumference; the line joining the two points of intersection, will pass through the same point.

Let the two sides AD, BC, of the trapezium ABCD inscribed in the circle ABC, be produced, and let them meet in E; and from any point in AD produced, draw FH parallel to BE, meeting the side DC in H; and join FB, meeting the circumference in G; the line joining G, H will always pass through the same point.

F

H

Let GH produced meet the circle in I. Join AI, DG. The angle GDH=GBC in the same segment, and .. is equal to the alternate angle GFH; whence a circle may be described through the points G, H, D, F; and .. the angle DGH=DFH =DEB. But the angle DEB being always the same, DGI, and . DAI, and also the arc DI will be invariable, and D being a fixed point, I must be also; i. e. GH will always pass through I.

F F

(65.) If the diagonals of a quadrilateral figure inscribed in a circle cut each other at right angles, the rectangles contained by the opposite sides are together double of the quadrilateral figure.

Let ABCD be a quadrilateral figure inscribed in a circle, whose diagonals AC, BD cut each other at right angles in E; the rectangles contained by AB, CD, and AD, BC are together double of the figure.

A

B

D

E

For (Eucl. vi. D.) the rectangles contained by AB, CD, and AD, BC, are together equal to the rectangle AC, BD, i. e. to the rectangles contained by AC, BE and AC, ED. But the rectangle contained by AC, BE is double of the triangle ABC, and the rectangle contained by AC, ED, is double of ADC; hence the rectangles contained by AB, CD and AD, BC are together double of ABCD.

(66.) If a rectangular parallelogram be inscribed in a right-angled triangle, and they have the right angle common; the rectangle contained by the segments of the hypothenuse is equal to the sum of the rectangles contained by the segments of the sides about the right angle.

Let ABC be a right-angled triangle, in which the rectangular parallelogram DBEF is inscribed, having one of its angles at B; the rectangle AF, FC is equal to the rectangles AD, DB and BE, Draw EG perpendicular to FC. ADF, EFG being similar,

A

D

F G

EC together. The triangles

AD AF FG (EF=) BD,

.. the rectangle AD, DB is equal to the rectangle AF, FG. In the same manner,

AF: (FD=) EB :: EC : CG,

.. the rectangle BE, EC is equal to the rectangle AF, GC, whence the rectangles AD, DB and BE, EC are together equal to the rectangles AF, FG and AF, GC, i. e. to the rectangle AF, FC (Eucl. ii. 1.).

(67.) If on the diameter of a semicircle, two equal circles be described, and in the curvilinear space included by the three circumferences a circle be inscribed; its diameter will be to that of the equal circles in the proportion of two to three.

On AB the diameter of the semicircle ADB let two equal circles ACE, BCH be described; and in the curvilinear space let the circle DEG be inscribed; its diameter FG AC :: 2: 3.

F

G

E

H

A

B

I

C

Let O and I be the centres of the circles. which will pass through the point of contact E;

Join OI,

and pro

duce it to K. From C draw CD perpendicular to AB, which will pass through O. Then the rectangle KO, OE is equal to the square of OC;

and OE OC :: OC: OK,

.. OE : OC :: OE+OC : OC+OK :: CD : KE+

and OE CD: 1: 3,

[CD :: 1 : 2;

.. FG : (CD=) AC :: 2 : 3.

(68.) If through the middle point of any chord of a circle two chords be drawn; the lines joining their extremities will intersect the first chord at equal distances from the middle point.

Let ACB be a chord of the circle ABD, bisected in C; and let DCF, ECG be any chords drawn through C. Join DG, EF cutting AB in I and H; then will CI= CH.

D

E

A

H

B

C

K

P

Through H draw KHL parallel to DG, meeting DF in K, and GE produced in L. Because LH is parallel to-GI, the angle HLE=CG1=HFK, and the vertical angles at H are equal, .. the triangles LEH, HKF are equiangular,

:. LH : HE :: HF : HK,

and the rectangle LH, HK is equal to the rectangle HE, HF, i. e. to the rectangle AH, HB or the difference of the squares of AC and CH. The triangles CID, CHK may in like manner be proved to be equiangular, as also the triangles CHL, CIG; hence

KH: HC:: DI: IC,

and LH: HC :: GI : IC,

:. KH × LH : HC2 :: DI× IG : IC2.

But KH× LH= AC2 – HC1, and DI× IG=AC2 – IC2,

.. AC2 – HC2 : HC2 :: AC2 – IC2 : IC2

comp. AC HC :: AC: IC,

:. HC2= IC2, and HC=IC.

(69.) The longest side of a trapezium being given, and made the diameter of the circumscribed circle; also

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 O 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.

B

E

H

F

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