Let the point A be joined to the centre O and, through the same point, let I K be a chord perpendicular to A O; then IK is bisected

in the point A [I. 44 i]. Because AI is a mean proportional to A C and A G [50], and because A E is equal to A G [Const.], the line AI is also a mean proportional to A C and A E; therefore AI is equal to AF [xxii]. Because A H is to AI as AI is to A B [49] and because AF is equal to A I, the line AH is to AF as AF is to AB: but AD is also to A F as AF is to AB [45]; then A H is equal to A D [xvii]. Now, AB is to AC as AE is to AH [Dem.]; therefore A B is to AC as AE is to AD, which W. T. B. D.

COROLLARY. If the distances from the vertex of an angle to the intersections of the sides with two straight lines be reciprocally proportional, the four intersections are points of the same circumference (48).

THEOREM 51.

If two circumferences be tangent to one another, and if two secants pass through their point of tangence, the intercepts of each of them in the two circumferences are proportional to the intercepts of the other.

Let ACE and BDE be two circumferences tangent in E, and let A B and C D be two secants to both circumferences through the point H.

Let AC and BD be joined, and let IH be a tangent to the two circumferences at the point E [I. 52]. Because the angle ACE is equal to the angle A EI [29 ii], and the angle B D E equal to the angle BEH; because also the angle AEI is equal to its vertical B EH [3], the alternate angles ACE and BD E are equal to each other [i] and the lines AC and D B are parallel to one another [21]: therefore AE is to BE as CE is to DE [38 i], which W. T. B. D.

COROLLARY I. If an angle or vertical angles, be subtended by parallels, the circumference passing through the vertex and the extremities of each of the subtending lines, is tangent to that passing through the vertex and the extremities of another.

COROLLARY II. The intercepts of a common secant through the point of tangence of two tangent circumferences, are proportional to their radii.

Scholium. The intercepts of a common secant through the point of tangence of two equal tangent circumferences, are equal to each other (I. 72).

PROBLEMS.

PROBLEM 1.

At a point given in a given straight line, to construct an angle equal to a given angle.

Let X Y be the given straight line, A the point given in it and BCD the given angle.

From the centres C and A, with equal radii, describe two circumferences cutting CB, C D and A Y in E, F and H; join EF; from the centre H, with a radius equal to E F, describe a circumference cutting the circumference A in a point I, and join AI:. the angle HAI is the required angle.

Join HI; then HI is a chord of the circumference A. Because the circumferences C and A have equal radii [Const.], they are equal to each other [I. 38]. Because HI is a radius of the circumference H, the line HI is equal to EF [Const.]. But HI and E F are chords of the equal circumferences C and A; then the arcs HI and E F are

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equal to each other [I. 43]: therefore the centre angles HAI and ECF are equal to each other [26]; that is, the angle HAI is equal to the angle B C D, which was to be done.

PROBLEM 2. To find the sum of two, or any number of, given angles.

PROBLEM 3. To find the product of a given angle by a given. number.

PROBLEM 4. To find the difference of two given angles.

From the centre B, describe a circumference cutting the sides of the angle A B C in the points D and E; from the centres D and E, with a radius greater than half the distance of the points D and E, describe two circumferences secant in F, and draw a straight