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GEOMETRICAL EXERCISES ON BOOK III.

THEOREM I.

If AB, CD be chords of a circle at right angles to each other, prove that the sum of the arcs AČ, BD is equal to the sum of the arcs AD, BC. (Archimedis, Lemm. Prop. 9.)

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

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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.

PROBLEM I. 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 is 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 centre is C and whose diameter AB is produced to the given point D.

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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. Join DE and produce it to meet the circumference again in F

join also CE and CF.
Then in the triangle GDF, 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, (111. 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 centre C, draw CF perpendicular to AD meeting the circumference of the circle in F:

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 Ĝ is determined according to the required

conditions.

PROBLEM II.

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. (III. 33.)

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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 circumą ference 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.

THEOREM II.

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 centre and meeting the convex and concave circumferences, the convex is one third of the concave circumference. (Archimedis, Lemm. Prop. 8.)

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 centre D, and meeting the convex circumference in F, and the concave in E.

Then the arc BF is one third of the arc AE,

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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 centre 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.

THEOREM III.

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 centre of the circle, is a constant quantity,

B

E

D

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

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.

THEOREM IV.

If two chords of a circle intersect each other at right angles, the sum of the square described upon the four segments is equal to the square described upon the diameter. (Archimedis, Lemm. Prop. 11.)

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

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Draw the diameter AF, and join AC, AD, CF, DB. Then the angle ACF in a semicircle is a right angle, (111. 31.) and equal to the angle AED: also the angle ADC is equal to the

angle AFC. (III. 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, (111. 26.) and therefore also the chord DB is equal to the chord CF. (111. 29.)

Because AEC is a right-angled triangle, the squares of AE, EC are equal to the square of AC; (1. 47.) similarly, the squares of DE, EB are equal to the square of DB; therefore the squares of AE, EC, DE, EB, are equal to the squares of AC, DB;

but DB was proved equal to FC,
and the squares of AC, FC are equal to the square

of AF, wherefore the squares of AE, EC, DE, EB, are equal to the

square of AF, the diameter of the circle.

PROBLEMS.

3. Given the centre of a circle ; find its diameter by means of the compasses alone.

4. Through a given point within a circle, to draw a chord which shall be bisected in that point.

5. Through a point in a circle which is not the centre, to draw the least chord.

6. To draw that diameter of a given circle which shall pass at a given distance from a given point.

7. Draw through one of the points in which any two circles cut one another, a straight line which shall be terminated by their circumferences and bisected in their point of section.

8. Determine the distance of a point from the centre of a given circle, so that if tangents be drawn from it to the circle, the concave part of the circumference may be double of the convex.

9. Find two points in a given straight line from each of which if two tangents be drawn to two given points on the same side of the given circle, they shall make an angle equal to a given angle.

Is any limitation required?

10. Find a point without a given circle, such that the sum of the two lines drawn from it touching the circle, shall be equal to the line drawn from it through the centre to meet the circle.

11. From a given point without a circle, a straight line is drawn cutting a circle. Draw from the same point another line so as to intercept two arcs which together shall subtend an angle equal to a given angle.

12. `In a chord of a circle produced, it is required to find a point, from which if a straight line be drawn touching the circle, the line so drawn shall be equal to a given straight line. 13.

Determine the point without a circle, from which, if two straight lines be drawn touching the circle, they may form an equi: lateral triangle with the chord which joins the points of contact.

14. A, B, C, are three given points, find the position of a circle such that all the tangents to it drawn from the points A, B, C shall be equal to one another. What is that circle which is the superior limit to those that satisfy the above condition?

15. Two parallel chords in a circle are respectively six and eight inches in length, and are one inch apart; how many inches is the diameter in length ?

16. The radius of the circle ABDE whose centre is C, is equal to five inches. The distance of the line AB from the centre is four inches. The distance of the line DE from the centre is three inches. Required the lengths of the straight lines AB, DE.

17. Required the locus of the vertices of all triangles upon the same base, having the sum of the squares of their sides equal to a given square.

18. Draw a straight line which shall touch a given circle, and make a given angle with a given straight line.

19. Draw a straight line which shall touch two given circles : (1) on the same side; (2) on the alternate sides.

20. Through a given point without a given circle, draw a straight line which shall cut off a quadrantal arc of that circle.

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