Billeder på siden
PDF
ePub

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

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 verțically 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

H

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

Hence GF: FC: GK: KL,

and FC FH:: KL: KI,

.. GF: FH :: GK : KI;

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

Triangles

in or about circles 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.

From

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

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

A

B

For the angle ABD in a semicircle is a right angle; and ABC is equal to the sum of ABD and DBC in their difference in the other; and in each case DBC= DAC in the same segment.

one case; and is equal to

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

F

D

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

[blocks in formation]

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.

GG

Let DEF be a circle, inscribed in the right-angled triangle ABC. The difference between AC, and AB, BC, is equal to the diameter of the circle.

Find O the centre, and join OD, OE. Then the angles at D, B, and E being

B I

right angles, and OD=OE, OB is a square; and DB, BE are equal to OD, OE, i. e. are together equal to the diameter of the circle. Now (Eucl. iii. 36. Cor.) CE= CF, and AD=AF; i. e. AC is equal to AD and CE; whence it is less than the sides containing the right angle, by DB and BE, or by the diameter of the circle.

(4.) If a semicircle be inscribed in a right-angled triangle, so as to touch the hypothenuse and perpendicular, and from the extremity of its diameter a line be drawn through the point of contact to meet the perpendicular produced; the part produced will be equal to the perpendicular.

Let the semicircle ADE touch the hypothenuse BC of the right-angled triangle ABC in D, and the perpendicular in A; and from E let ED be drawn to meet AB produced in F; AB=BF.

B

A

E

Join AD. Since ADE is a right angle, ADF is also a right angle, and .'. equal to DAF, DFA together. But DAF is equal to BDA, since BD =BA, being tangents from the same point B without the circle; and .. the angle BFD= BDF, and BF= BD-BA.

« ForrigeFortsæt »