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. 26. cor.), and consequently AEFB is a semicircle; wherefore AEB is a right angle (III. 25.) 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. Take the angle ABE away from both, and the angle BAE remains equal to EBD. Again, the opposite angles BAE and BFE of the drilateral figure BAEF, being equal to two right angles (III. 21.), 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. C B D qua 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. PROP. XXX. 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. 6. cor.), join CA and draw DE perpendicular to CA (I. 5. cor.), from C with the distance CA describe a circle meeting DE in F, join CF cutting the interior circumference in G; AG being joined, is the tangent which was required. For the triangles ACG and FCD have the sides CA, CG equal to CF, A E F D CD, and the containing angle ACF common to both; they are, therefore, equal (I. 3.), and consequently the angle CGA is equal to CDF. But CDF is a right angle; whence CGA is likewise a right angle, and AG a tangent to the circle (III. 27.) Or thus. G On AC 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. 25.), and consequently AG, AK touch the circle DGHK at the points G, K (III. 26.) PROP. XXXI. 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 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 making an angle BAD equal to C (I. 4.), erect AE perpendicular to AD (I. 38.), draw EF to bisect AB at right angles (I. 5. cor.) 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 A B D\ F B C E right angles, the distance EA is equal to EB (III. 6.), and the circle described through A must also pass through the point B; and since EAD is a right angle, AD touches the circle at A (III. 29.), and the angle BAD, which was made equal to C, is equal to the angle in the alternate segment AGB (III. 28.) PROP. XXXII. THEOR. Two circles which meet in the straight line join ing their centres or its continuation, touch each other. Let the circles DCE, FCG meet at C in the direction of the straight F line which joins their centres A, B; they touch each other at that point. B I D E For draw BH to another point H in the circumference DCE. And because B is distinct from the centre A, the line BH is greater than BC (III. 8. cor. 2.), and consequently the point H lies without the circle FCG. Except, therefore, at the single point C, the circumference DCE does not meet FCG. BA E Cor. Hence a straight line extending through the centres of two circles will pass through their points of contact. PROP. XXXIII. 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, which must touch both circles. In the case of internal contact, the angle GAE is equal to ACE in the alternate segment, (III. 29.); and for the same reason, GAE or GAD is equal to ABD; consequently the angles ACE and ABD are equal, and therefore (I. 25.) the straight lines CE and BD are parallel. When the contact is external, the angle GAE is still equal to ACE, and its vertical D 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 is parallel to BD. B PROP. XXXIV. THEOR. If from any point in the diameter of a circle or its extension, straight lines be drawn to the ends of a parallel chord; the squares of these lines are together equivalent to the squares of the segments into which the diameter is divided. Let BEFD be a circle, BD its diameter produced, and A a point in this, from which the straight lines AE and AF are drawn to the ends of the parallel chord EF; the squares of AE and AF are together equivalent to the squares of AB and AD. For from the centre C, let fall the perpendicular CG upon AB (I. 6.), and join AG and CE. |