the two given points are both within the given

circle.

PROP. XCVI.

123. PROBLEM. To find a point in a straight line, given in position, from which if two straight lines be drawn to two given points, without the given line, they shall have, first, their difference, and, secondly, their aggregate, equal to a given finite straight line.

Let A, B be two given points, XY a straight

line of indefinite length, but given in position; and let C be a given finite straight line: It is required to find a point in XY, from which if two straight lines be drawn to A and B, they shall have, first, their difference equal to C.

From A draw (E. 11. 1.) AD 1 to XY; produce AD to E, and make DE-AD; from the centre B, at a distance equal to C, describe the

circle FG; also, describe (S. 95. 3.) a circle EAF, which shall pass through E and A, and which shall touch the circle GF, in F; let K be the centre of the circle EAF, which centre (E. 1. 3. cor.) is in XY: Then is K the point which was to be found.

For join K, A and K, B; .. (E. 11. 3. or E. 12 3.) KB passes through the point of contact F; and (E. 15. def. 1.) KA=KF; .'. KB — KA= BF; and (constr.) BF=C; .. KB — KA=C. And, by a like construction, may a point be found in XY, from which if two straight lines be drawn, to A and B, their aggregate shall be equal to a given straight line.

But, in this case, the two points A and E must fall within the circle described from the centre B, at a distance equal to that given line; otherwise, the problem is impossible.

124. COR. 1. Let AB be (E. 10. 1.) bisected in I, let (E. 12. 1.) KM be drawn to AB, and let the circumference EAF cut AB in A and R, and BK produced in H: Then it is manifest, (constr. and E. 3. 3.) that 2IM=BR; and (E. 36. 3. cor.) AB × BR=HB × BF; i. e. 2AB X IM=HB× BF, or HF × BF + BF* (E. 3. 2.)

Let now IN be taken in IM (S. 67. 3.) so that 2AB X IN=BF'; .., if 2AB X IN be taken from 2ABX IM, and if BF be taken from HFX BF+BF, there will remain 2ABXNM=HFX

BF or 2AKX BF; ... ABX NM=AKX BF. 125. COR. 2. There is only one point K, in YX, from which if straight lines be drawn to A and B, their difference shall be equal to the given line C.

PROP. XCVII.

126. PROBLEM. The base and the altitude of a triangle being given, together with the aggregate or the difference, of the two remaining sides, to construct the triangle.

Let BC be the given base of a ▲, and BE, drawn

1 to BC, equal to its given altitude: It is required to construct a ▲, which shall have BC for its base, its altitude equal to BE, and, first, the aggregate of its two remaining sides of a given length.

Through E draw (E. 31. 1.) EF parallel to BC; find (S. 96. 3.) a point A from which, if AB and AC be drawn, their aggregate shall be equal to the given aggregate; if, ., A, B and A, C bel

R

joined, it is manifest that ABC is the A, which was to be described.

And, in the same manner, by the help of S. 96. s., may the problem be solved if the difference, instead of the aggregate of the two sides of the A, be given.

PROP. XCVIII.

127, PROBLEM. Three points being given, to find a fourth, from which if straight lines be drawn to the other three, two of them shall be equal, and the difference between either of these and the third shall be equal to a given straight line.

Let A, B and L be three given points, and M

M

a given finite straight line: It is required to find a fourth point, from which, if three straight lines be drawn to L, A, and B, two of them shall be

equal, and the difference between either of these and the third shall be equal to M.

From L as a centre, at a distance equal to M, describe the circle CDE; describe (S. 95. 3.) a circle CAB, which shall pass through A and B, and which shall touch the circle CDE; and let K be the centre of the circle CAB: Then is K the point which was to be found.

For join K, L; .., (E. 11. 3. or E. 12. 3.) KL, produced, passes through the point C, in which the two circles CDE, CAB touch one another; join, also, K, A and K, B; .. (E. 15. def. 1.) KA, KB and KC are equal to one another; and KL≈ KC LC; but (constr.) LC=M;.. KL is equal to the difference between KC and M, that is, to the difference between KA, or KB and M.

128. PROBLEM.

PROP. XCIX.

To describe a circle that shall touch three given circles, of which two are equal to one another.

Let AB, CD, EF be three given circles, of which the two AB and CD are equal to one another: It is required to describe a circle which shall touch the three given circles AB, CD, and EF.