20. In a right-angled triangle, having given the sum of the base and hypothenuse, and the sum of the base and perpendicular; to construct the triangle.

21. Given the perimeter of a right-angled triangle whose sides are in geometrical progression; to construct the triangle.

22. Given the difference of the angles at the base, the ratio of the segments of the base made by the perpendicular, and the sum of the sides; to construct the triangle.

23. Given the difference of the angles at the base, the ratio of the sides, and the length of a third proportional to the difference of the segments of the base made by a perpendicular from the vertex and the shorter side; to construct the triangle.

24. Given the base of a right-angled triangle; to construct it, when parts, equal to given lines, being cut off from the hypothenuse and perpendicular, the remainders have a given ratio.

25. Given one angle of a triangle, and the sums of each of the sides containing it and the third side; to construct the triangle.

26. Given the vertical angle, and the ratio of the sides containing it, as also the diameter of the circumscribing circle; to construct the triangle.

27. Given the vertical angle, and the radii of the inscribed and circumscribing circles; to construct the triangle.

28. Given the vertical angle, the radius of the inscribed circle, and the rectangle contained by the straight lines drawn from the centre of that circle to the angles at the base; to construct the triangle.

29. Given the base, one of the angles at the base, and the point in which the diameter of the circumscribing circle drawn from the vertex meets the base; to construct the triangle.

30. Given the vertical angle, the base, and the difference between two lines drawn from the centre of the inscribed circle to the angles at the base; to construct the triangle.

31. Given that segment of the line bisecting the vertical angle which is intercepted by perpendiculars let fall upon it from the angles at the base; the ratio of the sides; and the ratio of the radius of the

inscribed circle to the segment of the base which is intercepted between the line bisecting the vertical angle and the point of contact of the inscribed circle; to construct the triangle.

32. Given the line bisecting the vertical angle, and the differences between each side and the adjacent segment of the base made by the bisecting line; to construct the triangle.

33. Given one of the angles at the base, the side opposite to it, and the rectangle contained by the base and that segment of it made by the perpendicular which is adjacent to the given angle; to construct the triangle.

34. Given the vertical angle, and the lengths of two lines drawn from the extremities of the base to the points of bisection of the sides; to construct the triangle.

35. Given the lengths of three lines drawn from the angles to the points of bisection of the opposite sides; to construct the triangle.

36. Given the segments of the base made by the perpendicular, and one of the angles at the base triple the other; to construct the triangle.

37. The area and hypothenuse of a right-angled triangle being given; to construct the triangle.

38. Given one angle, and a line drawn from one of the others bisecting the side opposite to it; to construct the triangle, when the area is also given.

39. In two similar right-angled triangles, the sum of the base of one and perpendicular of the other is given; to determine the triangles such that their hypothenuses may contain the right angle of another triangle similar to them, and the sum of the three areas may be equal to a given area.

40. Given the vertical angle, the area, and the distance between the centres of the inscribed circle and the circle which touches the base and the two sides produced; to construct the triangle.

41. Given the area, the line from the vertex dividing the base into segments which have a given ratio, and either of the angles at the base; to construct the triangle.

42. Given the difference between the segments of the base made by the perpendicular, the sum of the squares of the sides, and the area; to construct the triangle.

43. Given the base, one of the angles at the base, and the difference between the side opposite to it and the perpendicular; to construct the triangle.

44. Given the vertical angle, the difference of the base and one side, and the sum of the perpendicular drawn from the angle at the base contiguous to that side upon the opposite side and the segment cut off by it from that opposite side contiguous to the other angle at the base; to construct the triangle.

45. Given the base, the difference of the sides, and the segment intercepted between the vertex and a perpendicular from one of the angles at the base upon the opposite side; to construct the triangle.

46. Given the vertical angle, the side of the inscribed square, and the rectangle contained by one side and its segment adjacent to the base made by the angular point of the inscribed square; to construct the triangle.

GEOMETRICAL

PROBLEMS.

SECT. I.

(1.) FROM a given point, to draw the shortest line possible to a given straight line.

Let A be the given point, and BD the given line. From A let fall the perpendicular AC; this will be less than any other line AD drawn from A to BD.

B

C

D

For since AC is perpendicular to BD, the angle ACD is a right angle, therefore the angle ADC is less than a right angle (Eucl. i. 32.) and consequently less than ACD. But the greater angle is subtended by the greater side (Eucl. i. 19.); therefore AD is greater than AC. In the same manner every other line drawn from A to BD may be shewn to be greater than AC; therefore AC is the least.

(2.) If a perpendicular be drawn bisecting a given straight line; any point in this perpendicular is at equal

A

GEOMETRICAL PROыLems.

[Sect. 1.

distances, and any point without the perpendicular is at unequal distances from the extremities of the line.

From C the point of bisection let CD be drawn at right angles to AB; any point D is at equal distances from

A and B DB. Since AC CB

Join AD, =

and CD is common, and the angle ĶD=BCD being right angles, AD

A

D

= DB. And the same may be proved of lines drawn from any other point in CD to A and B.

But if a point E be taken which is not in CD, join EA cutting the perpendicular in D; join EB, DB. Then AD=DB from the first part, and AE is equal to AD, DE, that is, to BD, DE, and is therefore greater than BE, (Eucl. i. 20.); therefore, &c.

(3.) Through a given point to draw a straight line which shall make equal angles with two straight lines given in position.

Let P be the given point, and BE, CF the lines given in position. Produce BE, CF to meet in A, and bisect the angle BAC by the line

B

E

D

P

AD. From P let fall the perpendicular PD, and produce it both ways to E and F. It will be the line required.

For the angle EAD is equal to the angle FAD, the angles at D right angles, and AD common, therefore (Eucl. i. 26.) the angle AED is equal to the angle AFD; therefore, &c.