Cor. If the given angle is a right angle, the figure will be a rectangle; and if, at the same time, the sides are equal, it will be a square. PROBLEM XIII. To find the center of a given circle or arc. Let ABC be the given circle or arc; it is required to find its center. Take any three points in the arc, as A, B, C, and join AB, BC. Bisect AB in D (Prob. I.), and through D draw DF A perpendicular to AB (Prob. II.). In the same manner, draw EF perpendicular to BC at its middle point. The perpendiculars DF, EF will meet in a point F equally distant from the points A, B, and C (Prop. VII., B. III.); and therefore F is the center of the circle. B Scholium. By the same construction, a circumference may be made to pass through three given points A, B, C; and also, a circle may be described about a triangle. PROBLEM XIV. Through a given point, to draw a tangent to a given circle. First. Let the given point A be without the circle BDE; it is required to draw a tangent to the circle through the point A. Find the center of the circle C, and join AC. Bisect AC in D; and with D as a center, and a radius equal to E C D AD, describe a circumference intersecting the given circumference in B. Draw AB, and it will be the tangent required. Draw the radius CB. The angle ABC, being inscribed in a semicircle, is a right angle (Prop. XV., Cor. 2, B. III.). Hence the line AB is a perpendicular at the extremity of the radius CB; it is, therefore, a tangent to the circumference (Prop. IX., B. III.). Secondly. If the given point is in the circumference of the circle, as the point B, draw the radius BC, and make BA perpendicular to BC. BA will be the tangent required (Prop. IX., B. III.). Scholium. When the point A lies without the circle, two tangents may always be drawn; for the circumference whose center is D intersects the given circumference in two points. PROBLEM XV. To inscribe a circle in a given triangle. Let ABC be the given triangle; it is required to inscribe a circle in it. E A F Bisect the angles B and C by the lines BD, CD, meeting each other in the point D. From the point of intersection, let fall the perpendiculars DE, DF, DG on the three sides of the triangle; these perpendiculars will all be equal. For, by construction, the angle B EBD is equal to the angle FBD; the right angle DEB is equal to the right angle DFB; hence the third angle BDE is equal to the third angle BDF (Prop. XXVII., Čor. 2, B. 1.). Moreover, the side BD is common to the two triangles BDE, BDF, and the angles adjacent to the common side are equal; therefore the two triangles are equal, and DE is equal to DF. For the same reason, DG is equal to DF. Therefore the three straight lines DE, DF, DG are equal to each other; and if a circumference be described from the center D, with a radius equal to DE, it will pass through the extremities of the lines DF, DG. It will also touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right angles (Prop. IX., B. III.). Therefore the circle EFG is inscribed in the triangle ABC (Def. 11, B. III.) Scholium. The three lines which bisect the angles of a triangle, all meet in the same point, viz., the center of the inscribed circle. PROBLEM XVI. Upon a given straight line, to describe a segment of a circle which shall contain a given angle. Let AB be the given straight line, upon which it is required to describe a segment of a circle containing a given angle. At the point A, in the straight line AB, make the angle BAD equal to the given angle; and from the point A draw AC perpendicular to AD. Bisect AB in E, and from E draw EC perpendicular to AB. From the point C, where these perpendiculars meet, with a radius equal to AC, describe a circle. Then will AGB be the segment required. For, since AD is a perpendicular at the extremity of the radius AC, it is a tangent (Prop. IX., B. III.); and the angle BAD is measured by half the arc AFB (Prop. XVI., B. III.). Also, the angle AGB, being an inscribed angle, is measured by half the same arc AFB; hence the angle AGB is equal to the angle BAD, which, by construction, is equal to the given angle. Therefore all the angles inscribed in the segment AGB are equal to the given angle. Scholium. If the given angle was a right angle, the required segment would be a semicircle, described on AB as a diameter. PROBLEM XVII. To divide a given straight line into any number of equal parts, or into parts proportional to given lines. First. Let AB be the given straight line which it is proposed to divide into any number of equal parts, as, for example, five. From the point A draw the indefinite straight line AC, making any angle D with AB. In AC take any point D, A E B and set off AD five times upon AC. Join BC, and draw DE parallel to it; then is AE the fifth part of AB. For, since ED is parallel to BC, AE: AB:: AD: AC (Prop. XVI., B. IV.). But AD is the fifth part of AC; therefore AE is the fifth part of AB. Secondly. Let AB be the given straight line, and AC a divided line; it is required to divide AB similarly to AC. Suppose AC to be divided in the points D and E. Place AB, AC so as to contain any angle; join BC, and through the points D, E draw DF, EG parallel to BC. The line AB will be divided into parts proportional to those of AC. For, because DF and EG are both parallel to CB, we have AD: AF:: DE: FG ::EC: GB (Prop. XVI., Cor. 2, B. IV.). PROBLEM XVIII. To find a fourth proportional to three given lines. From any point A draw two straight lines AD, AE, containing any angle DAE; and make AB, BD, AC respectively equal to the proposed lines. Join B, Č; and through D draw DE parallel to BC; then will CE be the fourth pro- A portional required. For, because BC is parallel to DE, we have AB: BD:: AC: CE (Prop. XVI., B. IV.). B D Cor. In the same manner may be found a third proportional to two given lines A and B; for this will be the same as a fourth proportional to the three lines A, B, B. PROBLEM XIX. To find a mean proportional between two given lines. Let AB, BC be the two given straight lines; it is required to find a mean proportional, between them. B C Place AB, BC in a straight line; upon AC describe the semicircle ADC; and from the point B draw BD perpendicular to AC. Then will BD be the mean proportional required. For the perpendicular BD, let fall from a point in the circumference upon the diameter, is a mean proportional between the two segments of the diameter AB, BC (Prop. XXII., Cor., B. IV.); and these segments are equal to the two given lines. 2 PROBLEM XX. To divide a given line into two parts, such that the greater part may be a mean proportional between the whole line and the other part. Let AB be the given straight line; it is required to divide it into two parts at the point F, such that AB: AF AF: FB. A B E At the extremity of the line AB, erect the perpendicular BC, and make it equal to the half of AB. From C as a center, with a radius equal to CB, describe a circle. Draw AC cutting the circumference in D; and make AF equal to AD. The line AB will be divided in the point F in the manner required. For, since AB is a perpendicular to the radius CB at its extremity, it is a tangent (Prop. IX., B. III.); and if we produce AC to E, we shall have AE: AB:: AB: AD (Prop. XXVIII., B. IV.). Therefore, by division (Prop. VII., B. II.), AE-AB : AB:: AB-AD: AD. But, by construction, AB is equal to DE; and therefore AE-AB is equal to AD or AF; and AB-AD is equal to FB. Hence AF: AB:: FB: AD or AF; and, consequently, by inversion (Prop. V., B. II.), AB: AF :: AF : FB. Scholium. The line AB is said to be divided in extreme and mean ratio. An example of its use may be seen in Prop. V., Book VI. PROBLEM XXI. Through a given point in a given angle, to draw a straight line so that the parts included between the point and the sides of the angle, may be equal. Let A be the given point, and BCD the given angle; it is required to draw through A a line BD, so that BA may be equal to AD. Through Through the point A draw AE parallel to BC; and take DE equal to CE. the points D and A draw the line will be the line required. A |