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6. A STRAIGHT LINE is said to be inscribed in a circle, when its extremities are in the circumference.

An inscribed angle is one which has its vertex in the circumference, and is included by two chords of the circle.

7. An inscribed triangle is one which has the vertices of its three angles in the circumference.

And generally, a polygon is said to be inscribed in a circle, when the vertices of all its angles are in the circumference. The circumference of the circle is then said to circumscribe the polygon.

8. A SECANT is a line which meets the circumference in two points, and lies partly within, and partly without the circle.

9. A TANGENT is a line which has but one point in common with the circumference.

The point where the tangent touches the circumference, is called the point of contact.

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10. Two circumferences touch each other when they have but one point in common. The common point is called the point of tangency.

11. A polygon is circumscribed about a circle, when each of its sides is tangent to the circumference. In the same case, the circle is said to be inscribed in the polygon.

POSTULATE.

12. Let it be granted that the circumference of a circle may be described from any centre, and with any radius.

PROPOSITION I. THEOREM.

Every diameter divides the circle and its circumference each into

two equal parts. Let AEBF, be a circle, and AB a diameter. Now, if the figure AEB be applied to AFB,

E their common base AB retaining its position, the curve line AEB must fall exactly on the curve line AFB, A A

B otherwise there would, in the one or the other, be points unequally distant from the centre, which is con

F trary to the definition of a circle. Hence, the diameter divides the circle and its circumference, each into two equal parts.

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Let AD be any chord. Draw the

D radii CA, CD, to its extremities. We shall then have (B. I., P. 7)*

А
AD<AC+CD,

B but AC plus CD is equal to AB; hence, AD<AB.

Cor. Hence, the greatest line which can be inscribed in a circle is a diameter.

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PROPOSITION III. THEOREM.

A straight line cannot meet the circumference of a circle in

more than two points. For, if it could meet it in three, those three points would be equally distant from the centre; and there would be three equal straight lines drawn from the same point to the same straight line, which is impossible (B. I., P. 15, c. 2).

* When reference is made from one Proposition to another, in the same Book, the number of the Proposition referred to is alone given; but when the Proposition is found in a different Book, the number of the Book is also given.

PROPOSITION IV. THEOREM.

In the same circle, or in equal circles, equal arcs are subtended by

equal chords: and conversely, equal chords subtend equal arcs.

Let C and O be the centres of two equal circles, and suppose the arc AMD equal to the arc ENG : then will the chord AD be equal to the chord EG.

For, since the diam. eters AB, EF, are equal, the semi-circle AMDB

M

N may be applied to the A

BE

F semi-circle ENGF, and the curve line AMDB

K will coincide with the curve line ENGF. But the part AMD is equal to the part ENG, by hypothesis; hence, the point D will fall on G; therefore, the chord AD will coincide with EG (B. I., A. 11), and hence, is equal to it (B. I., A. 14).

Conversely: If the chord AD is equal to the chord EG, the subtended arcs AMD, ENG, will also be equal.

For, drawing the radii CD, OG, the triangles A CD, EOG, will have their sides equal, each to each, namely, A0=EO, CD=OG, and AD=EG ; hence, the triangles are themselves equal; and, consequently, the angle ACD is equal to EOG (B. I., P. 10.)

Now, place the semi-circle ADB on its equal EGF, so that the radius AC may fall on the equal radius EO. Then, since the angle ACD is equal to the angle EOG, the radius CD will fall on OG, and the sector AMDC will coincide with the sector ENGO, and the arc AMD with the arc ENG : therefore, the arc AMD, is equal to the arc ENG (B. I., A. 14).

PROPOSITION V. THEOREM.

In equal circles, or in the same circle, a greater arc is subtend

ed by a greater chord: and conversely, the greater chord subtends the greater arc.

Let C be the common centre of two equal circles : then, if the arc ADH is greater than the arc AD, the chord AH will be greater than the chord AD.

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For, draw the radii CA, CD, OH, and the chords AD, AH. Now, the two sides AC, CH, of the triangle ACH are equal to the two sides AC, CD, of the triangle ACD, and the angle ACH, is greater than ACD: hence, the third side AH is greater than the third side AD (B. I., P. 9); there fore the chord which subtends the greater arc is the greater.

Conversely: If the chord AH is greater than AD, the arc ADH will be greater than the arc AD.

For, if ADH were equal to AD, the chord AH would be equal to the chord AD (P. 4), which is contrary to the hypothesis: and if the arc ADH were less than AD, the chord AH would be less than AD, which is also contrary to the hypothesis. Then, since the arc ADH, subtended by the greater chord, cannot be equal to, nor less than AD, it must be greater.

Scholium. The arcs here treated of are each less than the semi-circumference. If they were greater, the reverse property would have place; for, as the arcs increase, the chords will diminish, and conversely.

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The radius which is perpendicular to a chord, bisects the chord,

and bisects also the subtended arc of the chord.

Let AB be any chord, and CG a radius perpendicular to it: then will AD be equal to DB, and the arc AG to the arc GB. For, draw the radii CA, CB.

H Then the two right-angled triangles ADO, CDB, will have AC equal to CB, and CD common; hence, AD is equal to DB (B. I., P. 17).

A Again, since AD, DB, are equal, CG is a perpendicular

G

B

erected from the middle of AB; and since G is a point of this perpendicular, the chords AG and GB are equal (B. I., P. 16). But if the chord AG is equal to the chord GB, the arc AG is equal to the arc GB (P. 4); hence, the radius CG, at right angles to the chord AB, divides the arc subtended by that chord into two equal parts.

Scholium. The centre C, the middle point D of the chord AB, and the middle point G of the subtended arc, are three points of the same straight line perpendicular to the chord. But two points determine the position of a straight line (A. 11); hence, every straight line which passes through two of these points, will necessarily pass through the third, and be perpendicular to the chord.

It follows, also, that the perpendicular raised at the middle point of a chord passes through the centre of the circle, and through the middle point of the subtended arc.

For, the perpendicular to the chord, drawn from the centre of the circle, passes through the middle point of the chord, and only one perpendicular can be drawn from the same point to the same straight line (B. I., P. 14, c).

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Through three given points, not in the same straight line, one

circumference may always be made to pass, and but one.

Let A, B, and C, be the given points.

Join the points A and B by the straight line AB, and the points B and C by the straight line BC, and then bisect these

C

F lines by the perpendiculars DE

D FG: we say first, that DE and

A

K

B FG, will intersect in some point 0.

For, they intersect each other unless they are parallel (B. I., D. 16). Now, if they are

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