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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 Fis 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. .

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

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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 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 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.

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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 extermal, the angle GAE is still equal to ACE, and its vertical angle FAD is, for the same reason, equal to ABD; whence ACE is equal to ABD ; and these being alternate angles, the straight line CE (I. 22.) is parallel to BD.

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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. w

Because BF is parallel to CD, the - C arc BC is equal to the arc FD (III. A...T.so

18.), and consequently the chord BC / N #. \ is also equal to the chord FD (III. 12. \ |)

z

cor. 1.); but BC being the hypotenuse \
of the right-angled triangle BEC, its i.
square, or that of FD is equivalent to ~ 35

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

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therefore a right angle; consequently ACBF is a semicircle (III. 19, cor.) and AF the diameter. The angle ADF in the opposite semicircle is hence a right angle (III. 19.), and the square of the diameter AF is equal to the squares of AD and FD, or to the sum of the squares of the four segments EA, EB, ED, and EC intercepted between the circumference and the point E,

PROP. xxvi. THEOR. .

If through a point, within or without a circle, two straight lines be drawn to cut the circumference; the rectangle under the segments of the one, is equivalent to that contained by the segments of the other,

Let the two straight lines AD and AF be extended through the point A, to cut the circumference BFD of a circle; the rectangle contained by the segments AE and AF of the one, is equivalent to the rectangle under AB and AD, the distances intercepted from A in the other.

For draw AC to the centre, and produce it both ways to terminate in the circumference at G and H ; let fall the perpendicular CI upon BD (I. 6), and join CD.

Because CI is perpendicular to AD, the difference between the squares of CA and CD, the sides of the triangle ACD is equivalent to the difference between the squares of the segments AI and ID the segments of the base (II. 21. cor.); and the difference between the squares of two straight lines being equivalent to the rectangle under their sum and their

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