PROPOSITION XXXI. THEOREM.-If in a circle [1] an angle (BAC) be in a semicircle, it is a right angle; [2] but if the angle (ABC) be in a segment greater than a semicircle, it is less than a right angle; [3] and if the angle (ADC) be in a segment less than a semicircle, it is greater than a right angle CONSTRUCTION. Find the center E of the semicircle BAC (a), join AE, and produce BA to F. DEMONSTRATION. [1] Because in the triangle BAE the sides BE and AE are equal (b), the angles B and BAE are equal (c); and in the triangle EAC, because the sides EA and EC are equal (b), the angles EAC and ECA are equal (c); therefore the whole angle BAC is equal to the sum of the angles B and BCA. But the external angle FAC is also equal to the internal opposite angles B and BCA (d); therefore the angle FAC is equal to BAC, and they are each equal to a right angle (e). [2.] And because the two angles B and BAC of the triangle ABC are together less than two right angles (f), and BAC is a right angle, B must be less than a right angle; in a segment ABC greater than a semicircle angle. (a) III. 1. (b) I. Def. 15. (c) I. 5. (d) I. 32 A. (e) I. 13. (ƒ) I. 17. (g) III. 22. therefore the angle is less than a right [3.] In the segment ADC take any point D, and join AD and DC. Then because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal to two right angles (g); therefore B and D are together equal to two right angles. But B is less than a right angle; therefore D is greater than a right angle; and therefore the angle in the segment ADC less than a semicircle is greater than a right angle. SCHOLIA. 1. This proposition might be easily proved from the twentieth; for since the angle at the center of a semicircle is equal to two right angles, the angle at the circumference is equal to one right angle. And in like manner, in a segment greater or less than a semicircle, as the angle at the center is less than or greater than two right angles, so is the angle at the circumference less than or greater than one right angle. 2. The converse of this proposition is true, namely:-That if an angle in a segment is less than, equal to, or greater than a right angle, the segment is greater than, equal to, or less than a semicircle. If the hypotenuse of a right-angled triangle ABC be bisected in D, and from D as a center a circle be described with the radius AD, it will pass through each of the angles of the triangle A, B, and C. B COROLLARY 1. THEOREM. If a circle be described on the radius of another circle (ABC), and a straight line (AC) be drawn from the point (A) in which they meet to the outer circumference, that line will be bisected by the interior one. Join DE. Then because ADE is a semicircle, the angle ADE is a right angle (a); and because the straight line DE drawn from the center E is perpendicular to the chord AC, it bisects it (b). COROLLARY 2. THEOREM. If from a point (A), within or without a circle, two lines be drawn at right angles to each other, to meet the circumference, the sum of the squares on the segments (BA, CA, DA, and EA) between the point and the circumference is equal in area to the square on the diameter of the circle. Draw BF parallel to DE (a), and join CE, CF, EF, and BD. Then because BF and DE are parallel, the intercepted arcs DB and EF are equal (b), and therefore the chords DB and EF are also equal (c). Then in the right-angled triangle BAD the sum of the squares on AB and AD is equal in area to the square on BD, or to the square on its equal EF (d). And in the rightangled triangle AEC the sum of the squares on AE and AC is equal in area to the square on CE. Then because AC meets the two parallels, it makes the angles EAC and FBC equal (e); therefore FBC is a right angie, and the segment FDC a semicircle (f); therefore the angle FEC is a right angle (g). Then in the right-angled triangle FEC the sum of the squares on EF and EC is equal in area to the square on the diameter FC (d); but the sum of the squares on EF and EC is equal in area to the sum of the squares on AB, AD, AC, and AE; therefore the square on the diameter FC is equal in area to the sum of the squares on the segments AB, AC, AD, and AE. COROLLARY 3. PROBLEM. To draw a line through the extremity of a given line (AB), perpendicular to the same. From any point C without the given line, and with CB as a radius, describe a circle cutting AB in D; join DC, and produce the line to cut the circumference of the circle in E; draw BE, and it will be the perpendicular required. For the angle DBE, being in the semicircle EBD, is a right angle (a). (a) III. 31. COROLLARY 4. PROBLEM. To draw a tangent to a given circle (DBC), from a given point (A) without it. Find the center E.of the given circle (a), and join AE; upon AE, as a diameter, describe the circle AFD, cutting the circumference of the given circle in F and D; join AD and AF; then both AD and AF are tangents to the given circle. For join FE and DE; then because the line AE bisects the circle AFD, both the angles F and D are in semicircles, and are therefore right angles (b). From the solution of this proposition it is evident two tangents can be drawn from any point without a circle to the C 8 III. 1. same PROPOSITION XXXII. THEOREM.-If a straight line (EF) touches a circle, and from the point of contact (B) a straight line (BD) be drawn cutting the circle, the angles (EBD and FBD) formed by this line and the line touching the circle are equal to the angles (BCD and BAD) in the alternate segments of the circle. DEMONSTRATION. If the line BD is drawn at right angles to the line EF touching the circle, it passes through the center (a), and bisects the circle; therefore the angles A and C, being in the semicircles BAD and BCD, are right angles (6), and are equal to the angles EBD and FBD (c). E D E (a) III. 19. (b) III. 31. (c) Ax. 11. But if BD is not at right angles to EF, let AB be so drawn; then because ADCB is a semicircle, ADB is a right angle (6), and the other two angles A and DBA of the triangle DAB are together equal to a right angle (d), and therefore to the angle FBA, which is also a right E E B (b) III. 31. (d) I. 32 B, cor. 2. (ƒ) I. 13. angle. From these equals taking away the common angle DBA, the remaining angles A and FBD are equal. Again, in the quadrilateral figure ABCD, the opposite angles A and C are together equal to two right angles (e); but the angles EBD and FBD are also together equal to two right angles (f); therefore the angles A and C are equal to the angles EBD and FBD; and taking away the equals A and FBD, the remaining angles C and EBD are equal. SCHOLIUM. The first case considered above, namely, that in which the line BD is drawn at right angles to the touching line, has been usually omitted in the Elements. COROLLARY. THEOREM. If two straight lines (AB and AC) be drawn through the point of contact (A) of two circles, they intercept arcs, the chords of which (BC and DE) are parallel. FI F E (a) III. 32. (b) I. 28 A. (c) I. 15. (d) I. 27. Through the point of contact A draw the tangent FG. When one circle is within the other, the angles ADE and ABC are each equal to the angle CAG (a); therefore they are equal to each other; then because the straight line AB meets the two straight lines DE and BC, and forms the external angle ADE equal to the internal opposite angle ABC, the two lines DE and BC are parallel (b). Again, when each circle is without the other, because the angles CAG and EAF are vertical, they are equal (c); but the angle CAG is equal to B (a), and the angle D to EAF (a); therefore the angles B and D are equal; and because the straight line DB, meeting the two straight lines ED and BC, forms the alternate angles B and D equal the two lines ED and BC are parallel (d). PROPOSITION XXXIII. PROBLEM.-On a given finite straight line (AB) to describe a segment of a circle which shall contain an angle equal to a given rectilineal angle (C). Let the given angle be a right angle: SOLUTION. Bisect AB in F (a), and from the center F, with the radius AF, describe the semicircle AHB, and it will be the segment required. DEMONSTRATION. For the angle H in the semicircle AHB is a right angle (6), and the given angle is a right angle; in the segment AHB is equal to the given angle C. If the given angle be not a right angle: H E (a) I. 10. H therefore the angle Н B B D SOLUTION. At the point A in the straight line AB form the angle BAD equal to the given angle C (c), and from the point A draw AE perpendicular to AD (d); bisect AB in F (a), and from F draw FG perpendicular to AB (d); then from the center G, with the radius GA, describe the circle ABEI, and join GB and BE. AHB shall be upon the given line AB, and shall equal to the given angle C. (c) I. 23. Then the segment contain an angle DEMONSTRATION. Because in the triangles AFG and BFG the side AF is equal to BF, the side FG common to both, and the angle AFG equal to BFG, the base AG is equal to GB (e); therefore the circle AHE, described from G as a center, with the radius GA, shall pass through the point B; and because from the point A, at the extremity of the diameter AE, AD is drawn perpendicular to AE, therefore AD touches the circle (f); and because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB (g); but the |