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that each angle of ABCDE is equal to

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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. IX); hence,

ABF: abf AB2: ab2 (Theo. XIV).

=

Multiplying first couplet by 5,

=

5 ABF: 5 abf AB2: ab2.

But five times ABF is the area of ABCDE (Theo. XXIX, Book 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 is 1.72, what is the area of a regular pentagon whose side is 5?

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

5. If a quadrilateral is described about a circle, the angles subtended at the center by any two opposite sides are together equal to two right angles.

6. ABCD is a parallelogram, and P any point in the

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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 XVIII.

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 chords AC, BD, intersect each other in E. It is to be proved that AEX EC = BE × ED.

Join AD and BC. Now, in the triangles ADE, BCE, the vertical angles AED and BEC are equal (Theo. II, Book I);

D

also, the angles A and B are equal, being both measured by half the same arc, DC (Cor. 1, Theo. XXXIV, Book I);

hence, the third angles are equal, and the two triangles are similar (Theo. IX), and we have

A

AE: BEED: EC.

Then, multiplying extremes and means, we get

AE X EC BE × ED.

Therefore, if two chords, etc.

Cor. If through a fixed point within a circle any chord is

drawn, the product of its two segments is constant.

THEOREM XIX.

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

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Join BC and BD. Now, the angle DBA, contained by a tangent and chord, is measured by half the arc DB (Theo. XXXV, Book I); and the angle C, being an inscribed angle, is measured by half the same arc (Theo. XXXIV, Book 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. IX). Therefore,

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Cor. If from a point without a circle a secant is drawn, the product of the whole secant and its external part is constant.

Sch. As the distance from the center to a secant increases, the two points where it cuts the arc approach each other. When they coincide, the secant becomes a tangent.

THEOREM XX.

Similar arcs in different circles subtend equal angles at the centers.

Let AB and ab be similar arcs in two circles whose centers are C and c. It may be shown that the angles at C and c are equal.

B

a

Whatever portion AB is of the circumference to which it belongs, the angle C is the same portion of four right angles (Sch., Theo. XXXII, Book 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. XII). 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. XII). 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.

Sch. Since a degree is of the circumference in which it is taken (Sch., Theo. XXXII, Book I), it follows that degrees in unequal circumferences are similar but not equal arcs.

THEOREM XXI.

The circumferences of two circles are to each other as their diameters.

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the angles G and g, being subtended by similar arcs, AB and ab, are equal (Theo. XX); and the sides containing them are by construction proportional; therefore, the two triangles AGB, agb, are similar (Theo. XI), and we have

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But the perimeter ABCDEF is to the perimeter abcdef, as AB is to ab (Theo. XVI), and consequently as 2 AG is to 2ag. Now, if the number of sides of the similar polygons be indefinitely increased, the perimeters will ultimately coincide with the circumferences. Therefore,

Circumference ACE: circumference ace2 AG: 2ag. But 2 AG and 2 ag represent the diameters of the two circles (Def. 2, Sec. VIII, Book I).

Therefore, the circumferences, etc.

Cor. 1.-Representing the diameter of any circle by D, and its circumference by C, and remembering that the circumference of a circle whose diameter is 1 is 3.14159 (Sch., Theo. XLII, Book I), we have, by the above theorem,

1: D = 3.14159: C.

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