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Proposition 22. Theorem. 470. Of all polygons formed with the same given sides, that which can be inscribed in a circle is the maximum. Hyp. Let ABCDE be a

H

H' B' polygon inscribed in a•o, and A'B'C'D'E' other any

B polygon with the same sides as the first, but which can- E not be inscribed in a O.

To prove

ABCDE > A'B'C'D'E'.
Proof. Draw the diameter DH.
Join AH, HB.

Upon A'B' (= AB) cons. A A'B'H' = A ABH, and join D'H'.

Then, since the polygon HAED is inscribed in to with diameter HD .. HAED > H'A'E'D',

(469) and HBCD > H'B'C'D'.

(469) Adding, AHBCDE > A'H'B'C'D'E'. Subtracting, A ABH = AA'B'H'. ... ABCDE > A'B'C'D'E'.

(Ax. 5)

Q.E.D. 471. Cor. The maximum of all isoperimetric polygons of the same number of sides is regular. For, it is equilateral,

(468) and it can be inscribed in a 0,

(470) ... the polygon is regular.

(400)

EXERCISE.

Find a point in a given straight line such that the tangents drawn from it to a given circle contain the greatest angio possible.

Proposition 23. Theorem.

472. Of regular polygons with a given perimeter, that which has the greatest number of sides has the greatest area.

Hyp. Let P be a regular polygon of three sides, and Q a regular polygon of four sides, with the

Q same given perimeter.

Q > P.

To prove

Proof. In any side AB of P take any pt. C.

The polygon P may be regarded as an irregular polygon of four sides, in which the sides AC, CB make with each other a st. L.

Then, since the polygons P and Q are isoperimetric, and have the same number of sides,

(Hyp.)

... the irregular polygon P < the regular polygon Q. (471)

In the same way it may be shown that Q < the regular isoperimetric polygon of five sides, and so on. Q.E.D.

473. COR. The circle has a greater area than any polygon of the same perimeter.

EXERCISES.

1. Of all triangles of given base and area, the isosceles is that which has the greatest vertical angle.

2. The shortest chord which can be drawn through a given point within a circle is the perpendicular to the diameter which passes through that point.

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Proposition 24. Theorem. 474. Of regular polygons having the same given area, the greater the number of sides the less will be the perimeter.

Hyp. Let P and Q be regular polygons having the same area, and let Q have the greater number P. of sides.

To prove the perimeter of P> that of Q.

R
Proof. Let R be a regular poly-
gon having the same perimeter as
Q and the same number of sides as P.

Then, since Q has the same perimeter as R and a greater number of sides,

(Cons.) .'. > R.

(472) But Q=P,

(Hyp.) ...P> R. ... the perimeter of P > the perimeter of R. (379)

... the perimeter of P > the perimeter of Q. Q.E.D. 475. COR. The circumference of a circle is less than the perimeter of any polygon of the same area.

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Proposition 25. Theorem. 476. Given two intersecting straight lines AB, AC, and a point p between them ; then of all straight lines which pass through P and are terminated by AB, AC, that which is bisected at P cuts off the triangle of minimum area.

Hyp. Let EF be the st. line, terminated by AB, AC, which is bisected

F at P.

H
To prove A AEF is a minimum.
Proof. Let HK be any other st.

B line passing through P.

a

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.:. A AEF <A AKH.

(Ax. 4) In the same way it may be shown that A AEF < any other a formed by a st. line through P. .. A AEF is a minimum.

Q.E.D.

a

Proposition 26. Problem. 477. To find at what point in a given straight line the angle subtended by the line joining two given points, which are on the same side of the given straight line, is a maximum.

Given, the st. line CD, and the pts. A, B, on the same side of CD.

Required, to find at what pt. in CD the / subtended by the st. line AB is a maximum.

D

A

Cons. Describe a o to pass through
A, B, and to touch the st. line CD. (339) and Ex. 40 in (354)

Let P be the pt. of contact.
Then the Z APB is the required max. L.
Proof. Take any other pt. in CD as Q, and join AQ, BQ.

Then
ZAQB < ZAPB.

(246) ..ZAPB > any other / subtended by AB at a pt. in CD.

Q.E.F.

As

Proposition 27. Problem. 478. In a straight line of indefinite length find a point such that the sum of its distances from two given points, on the same side of the given line, shall be a minimum.

Given, the st. line CD, and the pts. A, B, on the same side of CD. Required, to find a pt. P in C

-D CD, so that the sum of AP, PB is a minimum.

Cons. Draw AF I to CD;
and produce AF to E, making FE = AF.

Join EB, cutting CD at P.
Join AP, PB.

Then P is the required pt., and of all lines drawn from A and B to a pt. in CD, the sum of AP, PB is a minimum.

Proof. Let Q be any other pt. in CD.
Join AQ, BQ, EQ.
Then
rt. A AFP = rt. A EFP.

(104)

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.. AP + PB < AQ + QB.
... the sum of AP and PB is a minimum.

a

Q.E.F.

479. COR. The sum of AP and PB is a minimum, when these lines are equally inclined to CD;

for

ZAPC = 2 EPC = LBPD.

NOTE.-In order that a ray of light from A may be reflected to a point B, it must fall upon a mirror CD at a point P where ZAPC = BPD ; i.e., by (478) the ray pursues the shortest path between A and B and touching CD.

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