cumference AHB, makes an acute angle with the diameter; but when it comes to meet the opposite semicircumference, it makes an obtuse angle. In passing, therefore, through all the intermediate gradations from minority to majority, the line DE must find a certain individual position in which it is at right angles to the diameter, and cuts the circle neither on the one side nor the other.

PROP. XXI. THEOR.

If, from the point of contact, a straight line be drawn to cut the circumference, the angles which it makes with the tangent are equal to those in the alternate segments of the circle.

Let CD be a tangent, and BE a straight line drawn from the point of contact, cutting the circle into two segments BAE and BFE; the angle EBD is equal to EAB, and the angle EBC to EFB.

For draw BA perpendicular to CD (I. 5. cor.), join AE, and from any point F in the opposite arc, draw FB and FE.

E

Because BA is perpendicular to the tangent at B, it is a diameter (III. 20. cor.), and consequently AEFB is a semicircle; wherefore AEB is a right angle (III. 19.), and the remaining acute angles BAE, ABE of the triangle, being together equal to another right angle, are equal to ABE and EBD, which compose the right angle ABD.

F

B

D

Take the angle ABE away from both, and the angle BAE

remains equal to EBD.

Again, the opposite angles BAE and BFE of the quadrilateral figure BAEF, being equal to two right angles (III. 17.), are equal to the angle EBD with its adjacent angle EBC; and taking away the equals BAE and EBD, there remains the angle BFE equal to EBC.

Cor. If a straight line meet the circumference of a circle, and make an angle with an inflected line equal to that in the alternate segment, it touches the circle.

Schol. A tangent may be considered as only a secant arrived at its ultimate position, when the two points through which it is drawn come to coincide. Suppose the straight line joining B and F were extended, it would make with the chord BE an angle EBF, equal to what the arc EF subtends from any point in the opposite circumference. But, when the point F is brought into the situation B, and BF merges into a tangent, the angle EBF passes into EBD, and the angle of the opposite or alternate segment becomes BAE.

PROP. XXII. PROB.

To draw a tangent to a circle, from a given point without it.

Let A be a given point, from which it is required to draw a straight line that shall touch the circle DGH.

Find the centre C(III. 5. cor.),

join AC, and on this as a diameter describe the circle AGCK, cutting the given circle in the points G, K: Join AG, AK; either of these lines is the tangent required.

H

For join CG, CK. And the angles CGA, CKA, be

ing each in a semicircle, are right angles (III. 19.), and consequently AG, AK, touch the circle DGHK at the points G, K (III. 20.).

Cor. Hence tangents drawn from the same point to a circle are equal; for the right angled triangles ACG and ACK having the side CG equal to CK, CA common, are equal (I. 21.), and consequently AG is equal to AK.

PROP. XXIII. PROB.

On a given straight line, to describe a segment of a circle, that shall contain an angle equal to a given angle.

Let AB be a straight line, on which it is required to describe a segment of a circle containing an angle equal to C.

If C be a right angle, it is evident that the problem will be performed, by describing a semicircle on AB. But if the angle C be either acute or obtuse; draw AD (I. 4.) making an angle BAD equal to C, erect AE (I. 34.), perpendicular A to AD, draw EF (I. 5. cor.) to bisect AB at right angles and meeting AE in E, and, from this point as a centre and with

the distance EA, describe the required segment AGB.

Because EF bisects AB at

G

E

B

F

D

G

F

C

E

right angles, the circle described through A must also pass through (III. 5.) the point B; and since EAD is a right

angle, AD touches the circle at A (III. 20.), and the angle BAD, which was made equal to C, is equal (III. 21.) to the angle in the alternate segment AGB.

PROP. XXIV. THEOR.

Two straight lines drawn through the point of contact of two circles, intercept arcs of which the chords are parallel.

Let the circles ACE and ABD touch mutually in A, and from this point the straight lines AC, AE be drawn to cut the circumferences; the chords CE and BD are parallel.

For draw the tangent FAG, (III. 20.), which must

touch both circles.

In the case of internal contact, the angle GAE is equal to ACE in the alternate segment, (III. 21.); and, for the same reason, GAE or GAD is equal to ABD; consequently the angles ACE and ABD are equal, and therefore (I. 22.) the straight lines CE and BD are parallel.

When the contact is external, the angle GAE is still e

qual to ACE, and its vertical

A

F

DA

angle FAD is, for the same reason, equal to ABD; whence

ACE is equal to ABD; and

B

E

these being alternate angles, the straight line CE (I. 22.)

is parallel to BD.

1

PROP. XXV. THEOR.

If through a point, within or without a circle, two perpendicular lines be drawn to meet the circumference, the squares of all the intercepted distances are together equivalent to the square of the diameter.

Let E be a point within or without the circle, and AB, CD two straight lines drawn through it at right angles to the circumference; the squares of the four segments EA, EB, ED, and EC, are together equivalent to the square of the diameter of the circle.

For draw BF parallel to CD, and join AF, AD, CB, and DF.

Because BF is parallel to CD, the arc BC is equal to the arc FD (III. 18.), and consequently the chord BC is also equal to the chord FD (III. 12. cor. 1.); but BC being the hypotenuse of the right-angled triangle BEC, its square, or that of FD is equivalent to the squares of EB and EC (II. 10.), and AED being likewise right-angled, the square of AD is equivalent to the squares of EA and ED. Whence the squares of AD and FD are equivalent to the four squares of EA, EB, ED, and EC. But since ED is parallel to

E

D

B

BE

F

BF, the interior angle ABF is equal to AED (I. 22.), and