AB, BE are given, and the contained angle ABE, which is the difference of ABC, EBC. The resolution of this triangle gives the angle BAE; and then, in the triangle ABD, the angles A and D, and the side AB are given ; whence AD, BD can be computed ; and CD may be found by means of the triangle ACD, or otherwise.

Schol. The solution is somewhat simplified, if A, B, C lie in a straight line: and still more, if they form a triangle, and D be in one of its sides. These cases present no difficulty.

It may be farther remarked, that if the circle BDCE pass through A, as it does when the angles BAC, BDC are supplementary, the problem is indeterminate or porismatic, since whereever the point D is taken in the arc BDC, the angles at D will be of the same magnitude.

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To describe a parallelogram having its adjacent sides respectively equal to the segments of one straight line divided in extreme and mean ratio, and its diagonals to those of another divided in the same manner.

Divide any straight line AB in extreme and mean ratio in C, and on AB describe the equilateral triangle ABD; through C draw CE, CF parallel to DB, DA : CEDF is a parallelogram such as is required.

For the triangles ACE, BCF are obviously equilateral, and CE, CF are therefore equal to the segments into which AB is divided in C. Again, BA: AC:: AC : CB, or DA : AC:: DF: DE, because BA is equal to DA, AC to DF, and CB to DE. Hence (VI. 6.) the triangles ADC, DFE are equiangular; and therefore (VI. 4. and alternately,) ED or CB : AC:: EF: CD, and, by composition, AB: AC:: CD + EF: CD. But, from the last analogy, by inversion, AC: CB, or (const.) AB : AC :: CD: EF; whence (V. 11.) CD + EF: CD::CD : EF; and therefore (VI. def. 3.) CD, EF are the segments of a line equal to their sum, cut in extreme and mean ratio.

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To draw a straight line cutting two given circles, so that the part of it within one of them may be equal to



a given straight line, and that it may divide the other into segments containing angles equal to given angles.

Let it be required to draw a straight line cutting the circles AB, CD, so that the part of it within AB may be equal to a given straight line, and that it may divide CD into segments containing angles equal to given angles.

In AB inscribe the chord BE equal to the given line, and (III. 34.) draw the straight line DF dividing CD into segments containing angles equal to the given angles ; from the centres G and H draw GK, HL perpendicular to the chords BE, DF, and with these £ perpendiculars as radii, and G, H as centres, describe circles; draw (App. I. 23.) MANC a common tangent to these circles ; it is the required line.

For (III. 14.) MA, BE are equal, because they are equally distant from the centre; and therefore MA is equal to the given line. For the same reason NC is equal to DF, and therefore (III. 28.) the arcs NC, DF are equal. But (III. 27.) in the same circle, the angles which stand on equal arcs are equal : therefore the segments NDFC and DNCF contain equal angles; and, since (const.) DNCF contains an angle equal to one of the given angles, and since MA is equal to the given straight line, MC is the line required.

Schol. MC may evidently have as many different positions, as the common tangent to the interior circles. It is plain also, that the given straight line must not exceed the diameter of the circle in which the part of the intersecting line equal to it is to be.



If there be two given unequal circles, which have not the same centre, two points may be found, through either of which if any straight line whatever be drawn cutting them, it will divide them into segments similar to one another, each to each.

Let ABC, DEF be the circles, and G, H their centres ; let the point K be so taken (VI. 10. and cor.) in the straight line passing through G and H, that its distances from those points may be proportional to the respective radii of the circles : any straight line drawn through K, and cutting the circles, cuts off similar segments.





Draw KFDCA cutting the circles, and join HF, HD; GC, GA. Then (const.) HK: HF:: GK: GC; wherefore (VÍ. 7.) the triangles KHF, KGC having the angle K common, and the angles KHF, KGC acute, are equiangular; and (I. 28.) HF, GC are parallel. In a similar manner it would be shown, that HD, GA are parallel. Hence (I. 34. cor. 4.) the angles FHD, CGA are equal ; and therefore the angles in the segments DEF, ABC being balves of these, are equal, and (III. def. 8.) these segments are similar; as are also the remaining segments, since (I. 22. and ax. 3.) the angles in them are equal.

Schol. 1. It is plain, (App. I. 24.) that when these points fall without the circles, the common tangents to the circles pass through them.

Schol. 2. If the circles be equal, there is one point which answers that which bisects the distance between the centres. The other may be regarded as lying at an infinite distance; and any parallel to the line joining the centres, which cuts the circles, divides them into proportional segments.


LET A be a point in the circumference of the circle ADG, and AB, BC, CD any number of equal arcs, and AE, EF, FG as many others equal to them, and all taken together less in amount than the circumference : then, if the chords AB, BE, CF, DG be drawn, and if AB, DG be produced to meet in H; GH is equal to the sum of the chords BE, CF, DG.

Draw ECK and FD. Then, because the arcs AB, BC, AE, &c., are equal, BE, CF, DG are parallel to one another; and so also are AH, EK, FD. Hence EH, FK are parallelograms, and HK, KD are equal to BE, CF. Add DG; then, HG is equal to BE, CF, DG.




From two given points to draw two straight lines meeting in a given straight line, and forming with one another the greatest angle possible.

From two given points A and B, let it be required to draw two

straight lines meeting in a given straight line CD, and making with one another the greatest angle possible.

Through the given points describe (App. II. 1.) a circle touching CD in D: D is the required point.

In CD take any point E on either side of D, and draw EB, EFA, and FB. Then (I. 16.) the exterior angle AFB is greater than AEB; and AFB is equal (111. 21.) to the angle contained by straight lines drawn from D to A and B; the last angle, therefore, is also greater than AEB; that is, than any angle formed by straight lines drawn from A and B, to any point in CD





except D.

Schol. 1. If A and B be at unequal distances from CD, there may be two circles touching the given line in D and D'; and the point D gives the maximum angle of those on one side of the straight line ABC, while D' is the vertex of the maximum angle among those on the other side of ABC. This will receive illustration from supposing straight lines drawn from A and B to a point varying its position along EC ; since the angle contained by those lines will continually diminish, as the point moves from D to C, where it vanishes. The point then continuing to move on towards D', the angle gradually increases, till the point arrives at D', when the angle is again a maximum, and beyond that point it continually diminishes.

Schol. 2. Of the angles formed by straight lines drawn from two given points to meet in the circumference of a given circle, those are the greatest or least possible which have their vertices at a point of contact of a circle passing through the given points, and touching the given circle.


In a triangle ABC, each angle of which is less than a right angle and a third, to find a point such that the sum of the straight lines drawn from it to the three angles may be a minimum.

On two of the sides, BC, CA, describe (III. 33.) segments of circles containing each an angle equal to a right angle and a third ; from D, their point of intersection, draw DA, DB, DC: these are the lines required.

Since ADC, BDC are each a right angle and a third, and the three angles at D are equal to four right angles, ADB must also


be of the same magnitude. Now, if AD be supposed to be of a fixed magnitude, and with it as radius, and A as centre, a circle be described, the radius AD makes equal angles with DB, DC; and therefore (App. I. 13. cor.) the sum of BD, DC is a minimum. Hence it is plain that the sum of the three lines cannot be a minimum, unless the angles ADB, ADC be equal. In the same manner it would be proved, that if BD be assumed of a fixed magnitude, the sum of AD, DC, is a minimum; and that the sum of AD, BD, CD cannot be a minimum, unless BD make equal angles with DA, DC; that is, from what has been already proved, unless the three angles at D be equal. But they are so by construction ; and therefore D is the required point.

Method of Computation. It is plain that the centres of the arcs ADC, BDC, are the centres of equilateral triangles described externally on AC, BC; that is, they are the points K, H in the figure for the 26th proposition of the first book of this Appendix. Now by the demonstration of that proposition, the angle HKC is equal to FAC: and, since the sides FC, CA, and the contained angle ACF (=ACB + 60°) are given, the angle CAF or CKH, or in this proposition CEF, can be found by the second case of trigonometry. Then, since ACE is 30°, if a perpendicular be drawn from E to AC, EC can be computed; and thence in the right-angled triangle EGC, CG, which is half of CD, may be found.


Two opposite sides of a quadrilateral described about a circle are together equal to the other two.




Let ABCD be a quadrilateral, the sides of which touch the circle EG in E, F, G, H ; the sum of AB, DC is equal to the sum of AD, BC.

For (III. 17. schol.) AE is equal to AH, and BE to BF; whence, by addition, AB is equal to the sum of AH, BF. It would be proved in a similar manner, that DC is equal to the sum of HD, FC; whence again by addition, AB, DC are together equal to AD, BC.

Schol. The converse of this is also true. For, let the sum of AD, BC be equal to that of AB, DC, and (IV. 4.) describe a circle touching any three sides AB, AD, DC, in E, H, G; join


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