Therefore,

EI: CD :: AI : BD.

But, by hypothesis, EI: CD :: IO : DF.

Then (21),

AI: BD :: IO: DF.

Also, if we subtract the equal angles EIA and CDB from the equal angles EIO and CDF, the remainders AIO and BDF are equal. Hence, the triangles AIO and BDF are similar. In the same manner, prove that each of the triangles of the first polygon is similar to its corresponding triangle in the other. Therefore, the figures are similar (432).

As in the case of equal polygons (422 and 430), it is only necessary to the hypothesis of this proposition, that all the angles except three in one polygon be equal to the homologous angles in the other.

434. Theorem.-In similar polygons the ratio of two homologous lines is the same as of any other two homologous lines.

For, since the polygons are similar, the triangles which

E

B

G

compose them are also similar, and (309),

AE: BC: EI : CD :: AI : BD :: IO : DF, etc. This common ratio is the linear ratio of the two figures.

Let the student show that the perpendicular let fall from E upon OU, and the homologous line in the other polygon, have the linear ratio of the two figures.

435. Theorem.-The perimeters of similar polygons are to each other as any two homologous lines.

The student may demonstrate this theorem in the same manner as the corresponding propositions in triangles (312).

436. Theorem.-The area of any polygon is to the area of a similar polygon, as the square on any line of the first is to the square on the homologous line of the second. Let the polygons BCD, etc., and AEI, etc., be divided into triangles by homologous diagonals. The trian

D

area BCD: area AEI :: BD : AI :: area BDF: area

2

2

2

2

AIO :: BF: AO :: area BFG : area AOU :: BG : AU :: area BGH : area AUY.

Selecting from these equal ratios the triangles, area BCD area AEI :: area BDF : area AIO :: area BFG : area AOU :: area BGH : area AUY.

Therefore (23), area BCDFGHB : area AEIOUYA ::

area BCD : area AEI; or, as BC: AE; or, as the areas of any other homologous parts; or, as the squares of any other homologous lines.

437. Corollary. The superficial ratio of two similar polygons is always the second power of their linear

ratio.

EXERCISES.

438.-1. Compose two polygons of the same number of triangles respectively similar, but not similarly arranged.

2. To draw a triangle similar to a given triangle, but with double the area.

3. What is the relation between the areas of the equilateral triangles described on the three sides of a right angled triangle?

REGULAR POLYGONS.

439. A REGULAR POLYGON is one which has all its sides equal, and all its angles equal. The square and the equilateral triangle are regular polygons.

440. Theorem.-Within a regular polygon there is a point equally distant from the vertices of all the angles.

Let ABCD, etc., be a regular polygon, and let lines bisecting the angles A and B

extend till they meet at 0. These lines will meet, for the interior angles which they make with AB are both acute (137).

In the triangle ABO, the

angles at A and B are equal, being halves of the equal angles of the polygon. Therefore, the opposite sides AO and BO are equal (275).

Join OC. Now, the triangles ABO and BCO are equal, for they have the side AO of the first equal to BO of the second, the side AB equal to BC, because the polygon is regular, and the included angles OAB and OBC equal, since they are halves of angles of the polygon. Hence, BO is equal to OC.

Then, the angle OCB is equal to OBC (268), and OC

bisects the angle BCD, which is equal to ABC. In the same manner, it is proved that OC is equal to OD, and so on. Therefore, the point O is equally distant from all the vertices.

CIRCUMSCRIBED AND INSCRIBED.

441. Corollary.-Every regular polygon may have a circle circumscribed about it. For, with O as a center and OA as a radius, a circumference may be described passing through all the vertices of the polygon (153).

442. Theorem.-The point which is equally distant from the vertices is also equally distant from the sides of a regular polygon.

BH C

The triangles OAB, OBC, etc., are all isosceles. If perpendiculars be let fall from O upon the several sides AB, BC, etc., these sides will be bisected (271). Then, the perpendiculars

will be equal, for they will be sides of equal triangles. But

they measure the distances from 0 to the several sides of the polygon. Therefore, the point O is equally distant from all the sides of the polygon.

443. Corollary.-Every regular polygon may have a circle inscribed in it. For with O as a center and OG as a radius, a circumference may be described passing through the feet of all these perpendiculars, and tangent to all the sides of the polygon (178), and therefore inscribed in it (253).

444. Corollary.-A regular polygon is a symmetrical figure.

445. The center of the circumscribed or inscribed circle is also called the center of a regular polygon. The

radius of the circumscribed circle is also called the radius of a regular polygon.

The APOTHEM of a regular polygon is the radius of the inscribed circle.

446. Theorem.-If the circumference of a circle be divided into equal arcs, the chords of those equal arcs will be the sides of a regular polygon.

For the sides are all equal, being the chords of equal arcs (185); and the angles are all equal, being inscribed in equal arcs (224).

447. Corollary.-An angle formed at the center of a regular polygon by lines from adjacent vertices, is an aliquot part of four right angles, being the quotient of four right angles divided by the number of the sides of the polygon.

448. Theorem.-If a circumference be divided into equal arcs, and lines tangent at the several points of division be produced until they meet, these tangents are the sides of a regular polygon.

Let A, B, C, etc., be points of division, and F, D, and E points where the tangents

meet.

Join GA, AB, and BC. Now, the triangles GAF, ABD, and BCE have the sides GA, AB, and BC equal, as they are chords of equal arcs; and the angles at G, A, B, and C equal, for each is formed by a tangent and chord which inter

F

D

cept equal arcs (226). Therefore, these triangles are all isosceles (275), and all equal (285); and the angles F, D, and E are equal. Also, FD and DE, being