ment; and shew that in the same circle, they are together equal to two right angles.

34. State and prove the converse of Euc. III. 22.

35. All circles which pass through two given points have their centers in a certain straight line.

36. Describe the circle of which a given segment is a part. Give Euclid's more simple method of solving the same problem independently of the magnitude of the given segment.

37. In the same circle equal straight lines cut off equal circumferences. If these straight lines have any point common to one another, it must not be in the circumference. Is the enunciation given complete? 38. Enunciate Euc. III. 31, and deduce the proof of it from Euc. 111. 20. 39. What is the locus of the vertices of all right-angled triangles which can be described upon the same hypotenuse?

40. How may a perpendicular be drawn to a given straight line from one of its extremities without producing the line?

41. If the angle in a semicircle be a right angle; what is the angle in a quadrant?

42. The sum of the squares of any two lines drawn from any point in a semicircle to the extremity of the diameter is constant. Express

that constant in terms of the radius.

43. In the demonstration of Euc. 111. 30, it is stated that "equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less:" explain by reference to the diagram the meaning of this statement.

44. How many circles may be described so as to pass through one, two, and three given points? In what case is it impossible for a circle to pass through three given points?

45. Compare the circumference of the segment (Euc. III. 33.) with the whole circumference when the angle contained in it is a right angle and a half.

46. Include the four cases of Euc. III. 35, in one general proot.

47. Enunciate the propositions which are converse to Props. 32, 35 of Book III.

48. If the position of the center of a circle be known with respect to a given point outside a circle, and the distance of the circumference to the point be ten inches: what is the length of the diameter of the circle, if a tangent drawn from the given point be fifteen inches?

49. If two straight lines be drawn from a point without a circle, and be both terminated by the concave part of the circumference, and if one of the lines pass through the center, and a portion of the other line intercepted by the circle, be equal to the radius: find the diameter of the circle, if the two lines meet the convex part of the circumference, a, b, units respectively from the given point.

50. Upon what propositions depends the demonstration of Euc. IIL 35? Is any extension made of this proposition in the Third Book?

51. What conditions must be fulfilled that a circle may pass through four given points?

52. Why is it considered necessary to demonstrate all the separate cases of Euc. III. 35, 36, geometrically, which are comprehended in one formula, when expressed by Algebraic symbols?

53. Enunciate the converse propositions of the Third Book of Euclid which are not demonstrated ex absurdo: and state the three methods which Euclid employs in the demonstration of converse propositions in the First and Third Books of the Elements.

PROPOSITION I. THEOREM.

If AB, CD be chords of a circle at right angles to each other, prove that the sum of the arcs AC, BD is equal to the sum of the arcs AD, BC.

Draw the diameter FGH parallel to AB, and cutting CD in H.

Then the arcs FDG and FCG are each half the circumference. Also since CD is bisected in the point H,

the arc FD is equal to the arc FC,

and the arc FD is equal to the arcs FA, AD, of which, AF is equal to BG,

therefore the arcs AD, BG are equal to the arc FC;

add to each CG,

therefore the arcs AD, BC are equal to the arcs FC, CG, which make up the half circumference.

Hence also the arcs AC, DB are equal to half the circumference. Wherefore the arcs AD, BC are equal to the arcs AC, DB.

PROPOSITION II. PROBLEM.

The diameter of a circle having been produced to a given point, it is required to find in the part produced a point, from which if a tangent be drawn to the circle, it shall be equal to the segment of the part produced, that is, between the given point and the point found.

Analysis. Let AEB be a circle whose center is C, and whose diameter AB is produced to the given point D.

Suppose that G is the point required, such that the segment GD is equal to the tangent GE drawn from G to touch the circle in E.

F

E

B G

Join DE and produce it to meet the circumference again in F; join also CE and CF.

Then in the triangle GDE, because GD is equal to GE,
therefore the angle GED is equal to the angle GDE;

and because CE is equal to CF,

the angle CEF is equal to the angle CFE;

therefore the angles CEF, GED are equal to the angles CFE, GDE:

but since GE is a tangent at E,

therefore the angle CEG is a right angle, (III. 18.)

hence the angles CEF, GEF are equal to a right angle, and consequently, the angles CFE, EDG are also equal to a right angle,

wherefore the remaining angle FCD of the triangle CFD is a right angle,

and therefore CF is perpendicular to AD. Synthesis. From the center C, draw CF perpendicular to AD meeting the circumference of the circle in F:

1

join DF cutting the circumference in E, join also CE, and at E draw EG perpendicular to CE and intersecting BD in G.

Then G will be the point required.

For in the triangle CFD, since FCD is a right angle, the angles CFD, CDF are together equal to a right angle;

also since CEG is a right angle,

therefore the angles CEF, GED are together equal to a right angle;

therefore the angles CEF, GED are equal to the angles CFD, CDF;

but because CE is equal to CF,

the angle CEF is equal to the angle CFD, wherefore the remaining angle GED is equal to the remaining angle CDF,

and the side GD is equal to the side GE of the triangle EGD, therefore the point G is determined according to the required conditions.

PROPOSITION III. THEOREM.

If a chord of a circle be produced till the part produced be equal to the radius, and if from its extremity a line be drawn through the center and meeting the convex and concave circumferences, the convex is one-third of the concave circumference.

Let AB any chord be produced to C, so that BC is equal to the radius of the circle:

and let CE be drawn from C through the center D, and meeting the convex circumference in F, and the concave in E. Then the arc BF' is one-third of the arc AE.

Draw EG parallel to AB, and join DB, DG.

Since the angle DEG is equal to the angle DGE; (1. 5.) and the angle GDF is equal to the angles DEG, DGE; (1. 32.) therefore the angle GDC is double of the angle DEG. But the angle BDC is equal to the angle BCD, (1. 5.) and the angle CEG is equal to the alternate angle ACE; (1. 29.) therefore the angle GDC is double of the angle CDB, add to these equals the angle CDB,

therefore the whole angle GDB is treble of the angle CDB, but the angles GDB, CDB at the center D, are subtended by the arcs BF, BG, of which BG is equal to AE.

Wherefore the circumference AE is treble of the circumference BF, and BF is one-third of AE.

Hence may be solved the following problem :

AE, BF are two arcs of a circle intercepted between a chord and a given diameter. Determine the position of the chord, so that one arc shall be triple of the other.

PROPOSITION IV. THEOREM.

AB, AC and ED are tangents to the circle CFB; at whatever point between C and B the tangent EFD is drawn, the three sides of the triangle AED are equal to twice AB or twice AC: also the angle subtended by the tangent EFD at the center of the circle, is a constant quantity.

Take G the center of the circle, and join GB, GE, GF, GD, GC. Then EB is equal to EF, and DC to DF; (111. 37.)

B E

C

therefore ED is equal to EB and DC;
to each of these add AE, AD,

wherefore AD, AE, ED are equal to AB, AC;
and AB is equal to AC,

therefore AD, AE, ED are equal to twice AB, or twice AC;
or the perimeter of the triangle AED is a constant quantity.
Again, the angle EGF is half of the angle BGF,
and the angle DGF is half of the angle CGF,
therefore the angle DGE is half of the angle CGB,

or the angle subtended by the tangent ED at G, is half of the angle contained between the two radii which meet the circle at the points where the two tangents AB, AC meet the circle.

PROPOSITION V. PROBLEM.

Given the base, the vertical angle, and the perpendicular in a plane triangle,

to construct it.

Upon the given base AB describe a segment of a circle containing an angle equal to the given angle. (11. 33.)

At the point B draw BC perpendicular to AB, and equal to the altitude of the triangle. (1. 11, 3.)

Through C, draw CDE parallel to AB, and meeting the circumference in D and E. (1. 31.)

Join DA, DB; also EA, EB;

then EAB or DAB is the triangle required.

It is also manifest, that if CDE touch the circle, there will be only one triangle which can be constructed on the base AB with the given altitude.

PROPOSITION VI. THEOREM.

If two chords of a circle intersect each other at right angles either within or without the circle, the sum of the squares described upon the four segments, is equal to the square described upon the diameter.

Let the chords AB, CD intersect at right angles in E.

Draw the diameter AF, and join AC, AD, CF, DB. Then the angle ACF in a semicircle is a right angle, (II. 31.) and equal to the angle AED:

also the angle ADC is equal to the angle AFC. (111. 21.) Hence in the triangles ADE, AFC, there are two angles in the one respectively equal to two angles in the other,

consequently, the third angle CAF is equal to the third angle DAB;

therefore the arc DB is equal to the arc CF, (II. 26.)

and therefore also the chord DB is equal to the chord CF. (III. 29.) Because AEC is a right-angled triangle,

the squares on AE, EC are equal to the square on AC; (1. 47.) similarly, the squares on DE, EB are equal to the square on DB; therefore the squares on AE, EC, DE, EB, are equal to the squares on AC, DB;

but DB was proved equal to FC,

and the squares on AC, FC are equal to the square on AF,