point and the points of tangency are equal, for the right triangles POA and POA' are equal (I. 83); therefore, PA == PA'. PROPOSITION XL.-PROBLEM. 92. To draw a common tangent to two given circles. Let O and O' be the centres of the given circles, and let the radius of the first be the greater. 1st. To draw an exterior common tangent. and a radius OM, equal to the difference of the given radii, describe a circumference; and from O' draw a tangent O'M to this circumference (90). Join OM, and produce it to meet the given circumference in A. Draw O'A' parallel to = With the centre 0, B B OA, and join AA'. Then AA' is a common tangent to the two given circles. For, by the construction, OMOA - O'A', and also OMOA - MA, whence MA O'A', and AMO'A' is a parallelogram (I. 108). But the angle M is a right angle; therefore, this parallelogram is a rectangle, and the angles at A and A' are right angles. Hence, AA' is a tangent to both circles. Since two tangents can be drawn from O' to the circle OM, there are two exterior common tangents to the given circles, namely, AA' and BB', which meet in a point T in the line of centres 00' produced. 2d. To draw an interior common tangent. With the centre 0 and a radius OM equal to the sum of the given radii, describe a circumference, and from O' draw a tangent O'M to this circumference. Join OM, intersecting the given circumference in A. Draw O'A' parallel to OA. Then, since OM = OA + O'A', we have AM O'A', and AMO'A' is a rectangle. Therefore, AA' is a tangent to both the given circles. There are two interior common tangents, AA' and BB', which intersect in a point T in the line of centres, between the two circles. 93. Scholium. If the given circles intersect each other, only the exterior tangents are possible. If they are tangent to each other externally, the two interior common tangents reduce to a single commcn tangent. If they are tangent internally, the two exterior tangents reduce to a single common tangent, and the interior tangents are not possible. If one circle is wholly within the other, there is no solution. PROPOSITION XLI.-PROBLEM. 94. To inscribe a circle in a given triangle. Α Let ABC be the given triangle. Bisect any two of its angles, as B and C, by straight lines meeting in O. From the point O let fall perpendiculars OD, OE, OF, upon the three sides of the triangle; these perpendiculars will be equal to each other (I. 129). Hence, the circumference of a circle, described with the centre O, and a radius = OD, will pass through the three points D, E, F, will be tangent to the three sides of the triangle at these points (26), and will therefore be inscribed in the triangle. B E 95. Scholium. If the sides of the triangle are produced and the exterior angles are bisected, the intersections O', O", O"", of the bisecting lines, will be the centres of three circles, each of which will touch one side of the triangle and the two other sides produced. In general, therefore, four circles can be drawn tangent to three intersecting straight lines. The three circles which lie without the triangle have been named escribed circles. PROPOSITION XLII.-PROBLEM. 96. Upon a given straight line, to describe a segment which shall contain a given angle. M Let AB be the given line. At the point B construct the angle ABC equal to the given angle. Draw BO perpendicular to BC, and DO perpendicular to AB at its middle point D, intersecting BO in. 0. With O as a centre, and radius OB describe the circumference AMBN. The segment AMB is the required segment. For, the line BC, being perpendicular to the radius OB, is a tangent to the circle; therefore, the angle ABC is measured by one-half the arc ANĖ (62), which is also the measure of any angle AMB inscribed in the segment AMB (57). Therefore, any angle inscribed in this segment is equal to the given angle. N B D C M 97. Scholium. If any point P is taken within the segment AMB, the angle APB is greater than the inscribed angle AMB (I. 74); and if any point Q is taken without this segment, but on the same side of the chord AB as the segment, the angle AQB is less than the inscribed angle AMB. Therefore, the angles whose vertices lie in the arc AMB are the only angles of the given magnitude whose sides pass through the two points A and B; hence, the arc AMB is the locus of the vertices of all the angles of the given magnitude whose sides pass through A and B. B' M' If any point M' be taken in the arc AM'B, the angle AMB is the supplement of the angle AM'B (61); and if BM' be produced to B', the angle AM'B' is also the supplement of AM'B; therefore AM'B' AMB. Hence the vertices of all the angles of the given magnitude whose sides, or sides produced, pass through A and B, lie B in the circumference AMBM'; that is, the locus of the vertices of all the angles of a given magnitude whose sides, or sides produced, pass through two fixed points, is a circumference passing through these points, and this locus may be constructed by the preceding problem. It may here be remarked, that in order to establish a certain line as a locus of points subject to certain given conditions, it is necessary not only to show that every point in that line satisfies the conditions, but also that no other points satisfy them; for the asserted locus must be the assemblage of all the points satisfying the given conditions (I. 40). INSCRIBED AND CIRCUMSCRIBED QUADRILATERALS. 98. Definition. An inscriptible quadrilateral is one which can be inscribed in a circle; that is, a circumference can be described passing through its four vertices. PROPOSITION XLIII.-THEOREM. 99. A quadrilateral is inscriptible if two opposite angles in it are supplements of each other. Let the angles A and C, of the quadrilateral ABCD, be supplements of each other. Describe a circumference passing through the three vertices B, C, D; and draw the chord BD. The angle A, being the supplement of C, is equal to any angle inscribed in the seg ment BMD (61); therefore the vertex A must B M Ꭺ D be or the arc BMD (97), and the quadrilateral is inscribed in the circle. 100. Scholium. This proposition is the converse of (61). PROPOSITION XLIV.-THEOREM. 101. In any circumscribed quadrilateral, the sum of two opposite sides is equal to the sum of the other two opposite sides. Let ABCD be circumscribed about a circle; For, let E, F, G, H, be the points of contact of the sides; then we have (91), E A H D AE AH, BE BF, CG = CF, DG = DH. Adding the corresponding members of these equalities, we have that is, AB + DC AD + BC. = G PROPOSITION XLV.-THEOREM. 102. Conversely, if the sum of two opposite sides of a quadrilateral is equal to the sum of the other two sides, the quadrilateral may be circumscribed about a circle. In the quadrilateral ABCD, let AB + DC= AD+BC; then, the quadrilateral can be circumscribed about a circle. B M Since the sum of the four angles of the quadrilateral is equal to four right angles, there must be two consecutive angles in it whose sum is not greater than two right angles; let B and C be these angles. Let a circle be described tangent to the three sides AB, BC, CD, the centre of this circle being the intersection of the bisectors of the angles B and C; then it is to be proved that this circle is tangent also to the fourth side AD. From the point A two tangents can be drawn to the circle (90). One of these tangents being AB, the other must be a line cutting CD (or CD produced); for, the sum of the angles B and C being not greater than two right angles, it is evident that no straight line |