Let the centre line O C be joined, and let the centres be joined to the points of intangence: then, because A B and D E are symmetric to the centre line [70], they will cut each other in a point H thereof [30i]. Because the circumferences O and C are equal to each other [Hyp.], AO is equal to BC [37 ii]; because AO and B C are both perpendicular to A B [55], they are parallel to each other [16]: therefore OC and A B bisect each other [25 i]; that is, H is the middle point of A B. The same reasoning would prove that H is also the middle point of DE; therefore AB and DE bisect each other on the centre line, W. W. T. B. D. Inversely Two circumferences the intangents of which cut each other in unequal parts, are not equal to each other. COROLLARY I. Conversely, two circumferences the intangents of which bisect each other, are equal to each other. COROLLARY II. The intangents of two equal circumferences bisect the centre line. COROLLARY III. The common secants through the points of intangence of two equal circumferences are parallel to each other; also they are perpendicular to the chords joining their extremities, or those of their outer parts, and conversely. COROLLARY IV. If from the middle point of the centre line of two equal circumferences, a tangent be drawn to one of them, its prolongation is tangent to the other. Scholium. Each intangent of two equal circumferences is bisected by the centre line. THEOREM 72. The intercepts of a common secant through the points of tangence of two equal tangent circumferences, are equal to each other. Let two equal circumferences OAH and CB H be tangent in H, and let A B be a common secant through the point H. Let OE and CD be two perpendiculars to AB from the centres O and C, and let OEM and CD N be two circumferences drawn con centric to the first with O E and CD as their respective radii; let also the centre line OC be joined: then AB is intangent to the circumferences O EM and CDN. Because OH is equal to CH [37ii], the two circumferences OEM and CDN are equal to each other [71 ii]; therefore O E is equal to CD. The chords A H and HB are then equally distant from the centres O and C [11]: therefore they are equal to each Inversely other [49 i], W. W.-T. B. D. If the intercepts of a common sécant through the point of tangence of two tangent circumferences be unequal, the circumferences are not equal to each other. COROLLARY I. If the intercepts of a common secant through the point of tangence of two tangent circumferences be equal to each other, the circumferences are equal to each other. COROLLARY II. The intercepts of a common secant through the middle point of the centre line of two equal circumferences are equal to each other, and conversely. COROLLARY III. The centre line of two equal circumferences bisects the common secant passing through its middle point, and conversely. COROLLARY IV. The continuation of the secant to one of two equa? circumferences through the middle point of the centre line, is secant to the other. THEOREM 73. The common secants which join the extremities of two common secants passing through the point of tangence of two equal tangent circumferences, are parallel and equal to each other. Let A CH and BDH be two equal circumferences tangent in H; let A B and C D be two secants through the point H, and let AD and C B be joined. A B Because the straight lines A B and C D are both bisected in H [72], they bisect each other; therefore A D and CB are parallel [26] and equal [26 i] to each other, W. W. T. B. D. COROLLARY I. The chords which join the extremities of two common secants passing through the point of tangence of two equal tangent circumferences, are parallel and equal to each other. COROLLARY II. The common secants, or the chords, which join the extremities of two common secants passing through the middle point of the centre line of two equal circumferences, or the extremities of their outer parts, are parallel and equal to each other. COROLLARY III. The common secants which join the extremities of two common secants passing through the point of tangence of two equal circumferences cut the intangent at equal distances from the point of tangence. THEOREM 74. If two parallel common secants to two equal tangent circumferences be equally distant from the point of tangence, the common secants joining their extremities cut each other in the point of tangence. Let A CE and B D E be two equal circumferences tangent in E, and let A B and C D be two parallel common secants equally distant from the point E. M A B N Let HI be the intangent of the two circumferences, cutting A B in H, and CD in I: then HE is equal to EI [23]. Let CE and DE be joined and produced to N and M, and let MN be joined. Because MN is parallel to CD [73] and passes through the point H [73 iii], it coincides with A B [17 iv]; thence the points M and N coincide respectively with the points A and B: that is, the continuation of DE passes through the point A, and that of CE, through the point B ; therefore the straight lines joining A to D and B to C cut each other in E, which W. T. B. D. COROLLARY I. If two parallel common secants to two equal tangent circumferences be equally distant from the point of tangence, they are equal to each other. COROLLARY II. The parallel common secants through the points of section of two equal secant circumferences, are equal to each other. PROBLEMS. The following problems are here solved as examples for students. The only instruments allowed for the solution of problems are the Ruler and Compasses; the Ruler being used for drawing and producing straight lines, the Compasses for describing circumferences and transferring dis tances. A PROBLEM 1. To find the sum of two given straight lines. Produce A B indefinitely and take, on the prolongation, B E equal to CD; the line A E is the required sum. Because B E is equal to CD, the sum of AB and BE is equal to that of A B and C D [ii]; thence A E is equal to the sum of AB and CD; therefore the sum of the given straight lines A B and C D has been found, which was to be done. PROBLEM 2. To find the sum of several straight lines. PROBLEM 3. To find the product of a given straight line by a given number. PROBLEM 4. To find the difference of two given straight lines. PROBLEM 5. Through a point given in a given straight line, to draw a perpendicular to that line. |