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(Dem.) For since ACBD is a quadrilateral inscribed in a circle, of which the diagonals are AB and CD, the rectangles AD CB and DB AC are together equal to the rectangle AB CD (VI. D).

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and

But the rectangle AD BC is equal to the rectangle AD AC, because BC is equal to AC; therefore the two rectangles ADAC and BD AC are equal to the rectangle ABCD.

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But the two rectangles AD AC and BD AC are the rectangle contained by AC, and the sum of the lines AD and DB (II. 1);

wherefore the rectangle contained by AC

and the sum of the lines AD and DB is equal to the rectangle AB⚫ CD; and because the sides of equal rectangles are reciprocally proportional (VI. 14), the sum of AD and DB is to DC as AB to AC.

PROPOSITION F. THEOREM.

If two points be taken in the diameter of a circle, such that the rectangle contained by the segments intercepted between them and the centre of the circle be equal to the square on the semidiameter; and if from these points two straight lines be inflected to any point whatsoever in the circumference of the circle, the ratio of the lines inflected will be the same with the ratio of the segments intercepted between the two first-mentioned points and the circumference of the circle.

Given a circle ABC, of which the centre is D, and in DA produced, let the points E and F be such that the rectangle ED DF is equal to the square on AD; from E and F to any point B in the circumference, let EB, FB be drawn; it is required to prove that FB: BE:: FA: AE.

(Const.) Join BD and BA,

(Dem.) and because the rectangle FD DE is equal to the square on AD, that is, on DB;

FD DB::DB: DE

(VI. 17). The two tri-
angles FDB and BDE have
therefore the sides propor-
tional that are about the
common angle D; there- F
fore they are equiangular
(VI. 6), the angle DEB
being equal to the angle
DBF,

and DBE to DFB.

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the angle DBA is equal but the angle DAB is equal to the and the angle DBA is equal to

Now, since DB is equal to DA, to the angle DAB (I. 5); angles AFB and FBA (I. 32),

the two angles ABE and EBD, therefore the angles AFB and FBA are equal to the angles ABE and EBD; but the angle AFB is equal to EBD, therefore the remaining angle FBA is equal ABE, wherefore the line AB bisects the vertical angle FBE of the triangle FBE; therefore (VI. 3) FB: BE :: FĂ: AE. COR.-If FB be produced to G, and BC joined; the two angles FBE and EBG are equal to two right angles (I. 13); but ABC, being an angle in a semicircle, is a right angle (III. 31); and the angle ABC is equal to the two angles ABE and EBC, therefore the two angles FBE and EBG are equal to twice the angles ABE and EBC; by the proposition the angle FBE is equal to twice the angle ABE, therefore the remaining angle EBG is equal to twice the angle EBC, wherefore the exterior angle EBG is bisected by the straight line BC; therefore (VI. A) FB: BE:: FC: CE, but FB: BE:: FA: AE, fore (V. 11) FC: CE :: FA: AE,

divided harmonically in A and E.

but

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and FC is therefore

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If, from one extremity of the diameter of a circle, a chord be drawn, and a perpendicular be drawn to the diameter, so as to cut it and the chord either internally or externally, the rectangle under the diameter and its segment by the perpendicular reckoned from that extremity, is equal to the rectangle under the chord and its corresponding segment.

Given a circle ABC, of which AC is a diameter, and DE a perpendicular to the diameter AC, and let AB meet DE in

B

E

F; to prove that
CA⚫ AD.

(Const.) Join BC.

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the rectangle BA AF is equal to the rectangle

(Dem.) And because ABC is an angle in a semicircle, it is a right angle (III. 31). Now, the angle ADF is also a right angle (Hyp.); and the same with DAF, or vertical to it; ABC, ADF are equiangular, and

angle BAC is either the therefore the triangles BA AC :: AD: AF

(VI..4); therefore also the rectangle BAAF, contained by the extremes, is equal to the rectangle AC AD, contained by the means (VI. 16).

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If the angles at the base of a triangle be bisected by two lines that meet, and the exterior angles at the base, formed by producing the two sides, be similarly bisected, the two points of concourse and the vertex shall be in one straight line, which shall bisect the vertical angle.

Given a triangle ABC, of which the angles at the base are bisected by the lines AD and BD, which meet in D, and the exterior angles are bisected by AG and BG; to prove that the two points D, G, and the vertex C, are in one straight line, which bisects the angle at C.

(Dem.) Then in the

(Const.) For draw DE, DF, DL, and GM, GK, GN perpendicular to the sides and sides produced. triangles ADE, ADL, the angles at A are equal, and the angles at E and I are also equal, being right angles, and AD is common to them; therefore (I. 26) AL = AE and DL = DE. It is proved in a similar manner and DL = DF; also that AM = AK, GM = GK;

that BL= BF,

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GK;

K

and

M

since DE and DF are each equal to DL, therefore DE = DF; and simi

larly GM GN.

=

Again,

since in

N

the triangles CED, CFD, the two sides CD and DE are respectively equal to CD and DF, and the angles at E and F are riglit, consequently (I. C.) the triangles are equal in every respect; and therefore the angle at C is bisected by CD. The triangles CMG and CNG are similarly proved (I. C.) to be equal in every respect; hence the angle at C is bisected by CG, but it was also shewn to be bisected by CD; therefore the lines CG, CD coincide; and therefore the three points C, D, and G are in the same straight line.

PROPOSITION K. THEOREM.

In reference to the triangle of proposition H, the segments of each side produced that are intercepted between the vertex and the external perpendiculars, are each equal to the semiperimeter of the triangle, the segments of these sides next the vertex are equal to the excess of the semiperimeter above the base, and the

M

segment of each of these sides next the base is respectively equal to the excess of the semiperimeter above the other side.

In the diagram of Prop. H. or L., CM or CN=S, if S denote the semiperimeter; CE or CFS - AB, AES BC, and BF

S-AC.

(Dem.) Since AK

= AM, and BK = BN, therefore CM + CN = perimeter, and CM CNS, the semiperimeter; = AMS AC, and BNS — BC.

=

and

Also CE+AE + AM = semiperimeter = CF+FB + BN, and CE = CF, therefore AE and AM are equal to FB and BN, or their equals AL + AK = BK + BL, that is 2AK + KL = 2BL + KL, take KL from both, then 2AK = 2BL or AK = = BL; these equals add KL, then AL = fore BK AE,

BK,

and AK AM,

=

to each of but AL = AE, therehence BK + KA = EA

+ AM, or AB = EM; therefore CE or CM - EM S― AB. Also BN = BK AL = AE = S - BC. =

Wherefore CM,

CE, EA, and AM are respectively equal to S, S — AB, S and S AC.

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PROPOSITION L. THEOREM.

The area of a triangle is a mean proportional between the rectangle under the semiperimeter and its excess above the base, and that under its respective excesses above the other two sides.

or

It is required to prove that the triangle ABC is a mean proportional between the rectangles CM CE, and AM·AE, between S(SAB) and (SAC) (S – BC).

(Dem.) For since the angle CED is equal to the angle CMG, each being a right angle, and

the angle MCG is common to the two

triangles CED and CMG,

they are

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therefore CE CM: ED. CM CM ED: MG ED (VI. 23, Cor. 1). But the triangle ABC is equal to the three ADC, ADB, and CDB, which are respectively equal to half the rectangles AC DE, AB DL, BC DF (I. 41), the sum of which is = :ED CM.

And since angles EAL + MAK EAD MAG = one right angle

=

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two right angles, AGM+MAG (I. 32), and therefore EAD = AGM, and the angles at E and M are right angles, hence the triangles AGM and EAD are similar, and AE: ED = MG: AM; therefore AE AM = MG ED. The former analogy therefore becomes CM CE

: ABC ABC : AM AE, (SAC) (SBC).

or S(SAB) : ABC = ABC:

COR.-If the sides opposite to the angles A, B, C, be respectively denoted by a, b, and C, then

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Which is the useful rule by which the area of a triangle is computed, when its three sides are given.

EXERCISES.

1. Lines that meet any three parallel lines, are cut proportionally.

2. If a straight line be divided into three segments such, that the rectangle under the whole line and the middle segment is equal to that under the extreme segments, the line is cut harmonically.

3. If from any point in the circumference of a circle a perpendicular be drawn on any radius, and a tangent from the same point to meet the radius produced, the radius will be a mean proportional between its segments, intercepted between the centre and the points of concourse.

4. If arcs of different circles have a common chord, lines diverging from one of its extremities will cut the arcs proportionally.

5. To cut a given line harmonically.

6. To find a line such, that the first of two given lines shall be to the second as the square of the first to the square of the required line.

7. To find a line such, that the first of two given lines shall be to it as the square of the first to the square of the second.

8. From one angle of a triangle a line is drawn to the point of bisection of the opposite side, and through its middle point another is drawn from either angle to the side subtending it; prove that this is divided in the ratio of 2 to 1.

9. If a square be inscribed in a right-angled triangle, one side being on the hypotenuse, the hypotenuse is divided in continued proportion.

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