For by similarity of the triangles BEF, BAD (Cor. Theo. V), if BE be one-half of BA, EF will be onehalf of AD; and in the same manner it may be shown that FG will be one-half of BC: consequently, the whole line EG will be one-half the sum of AD and BC.

Cor. 2. The area of the trapezoid (Theo. XVIII, B. I) is equal to its altitude multiplied by EG, which is its mean breadth.

THEOREM X.

The perimeters of similar polygons are to each other as their homologous sides.

Let

ABCDE,

abcde, be two simi

lar polygons, hav

ing the angles A,

B, C, etc., respectively equal to the

angles a, b, c, etc.

B

a

Then will their perimeters be to each other as any two homologous sides AB and ab.

By similarity of the two polygons (Def. 1, Sec. X) we have

AB: ab :: BC : bc :: CD: cd :: DE: de :: EA: ea.
And therefore, by composition (Def. 7, Sec. X),
AB+BC+CD+DE+EA: ab+bc+cd+de+ea ::

AB: ab.

But the first two terms of the last proportion represent the perimeters of the two polygons (Def. 3, Sec. VII, B. I.)

Therefore, the perimeters, etc.

THEOREM XI.

The areas of two regular polygons of the same number of sides are to each other as the squares of their sides.

former is to the area of the latter as AB2 is to ab2.

Since the two polygons have the same number of sides and angles, it is evident (Theo. XX, B. I) that the sum of all the angles of ABCDE is equal to the sum of all the angles of abcde, and consequently that each angle of ABCDE is equal to each angle of abcde (Def. 2, Sec. VII).

Bisect the angles A and B by AF and BF, and the angles a and b by af and bf. Now, since the triangles ABF, abf, have two angles of the one equal to two angles of the other, they are similar (Cor., Theo. III); hence,

ABF abf AB2: ab2 (Theo. VIII). Multiplying an extreme and a mean equally,

5 ABF: 5 abf:: AB2 : ab2 (Def. 5, Sec. X). But five times ABF is the area of ABCDE (Theo. XXI, B. I), and five times abf is the area of abcde.

Therefore, the areas, etc.

EXERCISES.

1. Prove that two parallelograms of the same altitude are to each other as their bases.

2. Prove that if two isosceles triangles have their vertical angles equal they are similar.

3. If the area of a regular pentagon whose side is 1 be 1.72, what is the area of a regular pentagon whose side is 5?

4. In two similar polygons, if a side of one be 7, and the homologous side of the other 13, and if the perimeter of the former be 39, what is the perimeter of the latter?

SEC. XI.-OF CIRCLES.

DEFINITIONS.

1. Two ARCS of circles are called similar when they are equal parts of the circumferences to which they belong.

2. Two SECTORS are called similar when their arcs are similar.

THEOREM XII.

If two chords intersect in a circle, the product of the parts of the one is equal to the product of the parts of the other.

In the circle ABCD let the D chords AC, BD, intersect each

C

other in E. It is to be proved

that AEXEC=BE>ED.

Join AD and BC. Now, in

the triangles ADE, BCE, the

E

B

vertical angles AED and BEC are equal (Theo. II,

B. I); also, the angles A and B are equal, being both
measured by half the same arc, DC (Theo. XXVI, B.
I); hence, the third angles are equal, and the two
triangles are similar (Theo. III), and we have
AE BEED: EC.

Then, multiplying extremes and means, we get
AEXEC=BEXED.

Therefore, if two chords, etc.

THEOREM XIII.

If from a point without a circle a tangent and a secant be drawn, the tangent will be a mean proportional between the secant and its external part.

Let AB be a tangent and AC a secant to a circle. Then will

AC AB: AB : AD.

Join BC and BD. Now, the angle DBA, contained by a tangent and chord, is measured by half the arc DB (Theo. XXVII, B. I); and the angle C, being an

B

2.11

inscribed angle, is measured by half the same arc (Theo. XXVI, B. I); therefore, these two angles are equal. But the angle A is common to the two triangles BAD and BAC. Hence, these triangles have two angles of the one equal to two angles of the other, and are consequently similar (Cor., Theo. III). Therefore, AC: AB:: AB: AD.

That is, if from a point, etc.

Whatever portion AB is of the circumference to which it belongs, the angle C is the same portion of four right angles (Schol., Theo. XXIV, B. I); and whatever portion ab is of the other circumference, the angle c is the same portion of four right angles. But AB is the same portion of the first circumference that ab is of the second (Def. 1, Sec. XI). Hence, the angles C and c are equal portions of four right angles; hence, also, they are equal to each other.

Therefore, similar arcs, etc.

Cor. The sectors ABC, abc, are similar (Def. 2, Sec. XI). Hence, the sides of similar sectors contain equal angles. Hence, also, similar sectors are contained an equal number of times in the circles to which they belong.

Schol. Since a degree is of the circumference in which it is taken (Schol., Theo. XXIV, B. I), it follows that degrees in unequal circles are similar but not equal arcs.