therefore the angle PCO is equal to the angle QCO;
and the angle ACB is bisected by CO:

also CO produced will bisect the arc AB in D.

(III. 26.)

If a tangent EDF be drawn to touch the arc AB in D;
and CA, CB be produced to meet it in E, F:

the inscription of the circle in the sector is reduced to the inscription of a circle in a triangle. (IV. 4.)

PROPOSITION III. THEOREM.

In any triangle inscribed in a circle, if from one extremity of that diameter which bisects the base, a perpendicular be drawn to the longer of the two sides; the segments of this side are respectively equal to half the sum and half the difference of the two sides of the triangle.

Let ABC be a triangle inscribed in a circle, of which the diameter DE bisects the base BC in F, and also the arc BEC in E.

Let EG be drawn from E, one extremity of the diameter DE, perpendicular on the longer side AB.

Then AG is equal to half the sum of the two sides AB, AC,
and BG equal to half their difference.

From E draw EH perpendicular on AC produced in H, and join AE, EB, EC.

First. Because the angles BAE, CAE are equal (Euc. III. 21.), and the angles EGA, EHA are right angles,

also the side AE is common to the two triangles AGE, AHE:
therefore EG is equal to EH, and AG to AH. (Euc. 1. 26.)
Again, because EB is equal to EC,

and EG to EH in the right-angled triangles EBG, ECH;
therefore BG is equal to CH.

And AG has been shewn to be equal to AH,
therefore AB and AC are equal to AG and AH;
but AG and AH are double of AG;

wherefore the double of AG is equal to AB and AC, and AG is equal to half the sum of AB and AC. Secondly. It has been shewn that AG is equal to AH, and BG to CH.

Because AB is equal to the sum of the lines AG, GB,

and AC equal to the difference of AH, CH;

therefore the difference of AB and AC is double of BG,
and BG is half the difference of AB and AC.

PROPOSITION IV. THEOREM.

In any triangle inscribed in a circle, if from the extremities of that diameter which bisects the base there be drawn two lines to the vertex of the triangle; the angle contained between the diameter and the line drawn to that extremity below the base, is equal to half the difference of the angles at the base of the triangle; and the angle contained by the line drawn to the other extremity and the adjacent side of the triangle, is equal to half their

sum.

Let ABC be a triangle inscribed in a circle whose diameter DE bisects the base BC, also the arc BEC in E; and let AE, AD be joined.

Then the angle AED is equal to half the difference of the angles ABC, ACB at the base of the triangle, and the angle BAD is equal to half their sum.

First. Draw AG parallel to BC, and join BG, GE.
Because BC, AG are parallel chords in the circle,
the arc BG is equal to the arc AC;
add to these equals the arc GDA,
therefore the arc BGÃ is equal to the arc CAG:
but the angle ACB stands on the arc BGA,
and the angle GBC on the arc CAG,
therefore the angle ACB is equal to the angle GBC;
and if ABC be taken from these equals,

the difference of the angles ACB, ABC is equal to GBA, the difference of the angles GBC, ABC,

but GBA is equal to GEA, (Euc. III.) or twice AED,

for ED bisects the angle GEA:

therefore twice the angle AED is equal to the difference of the angles ACB, ABC,

and the angle AED is equal to one-half the difference of ACB, ABC, the angles at the base of the triangle. Secondly. Let CA be produced to H.

Then because the arcs BE, EC are equal,
the angle BAE is equal to the angle EAC,
and AE bisects the angle BAC.

And because AD being at right angles to AE, (Euc. II. 31.) bisects the adjacent angle BAH,

therefore BAD is half of the angle BAH.

But BAH is equal to the two angles ABC, ACB; (Euc. 1. 32.) therefore BAD is equal to half the sum of the angles ACB, ABC.

PROPOSITION V. PROBLEM.

ABCD is a rectangular parallelogram. Required to draw EG, FG parallel to AD, DC, so that the rectangle EF may be equal to the figure EMD, and EB equal to FD.

Analysis. Let EG, FG be drawn, as required, bisecting the rectangle ABCD.

Draw the diagonal BD cutting EG in H and FG in K.

Then BD also bisects the rectangle ABCD;

and therefore the area of the triangle KGH is equal to that of the two triangles EHB, FKD.

Draw GL perpendicular to BD, and join GB,

also produce FG to M, and EG to N.

If the triangle LGH be supposed to be equal to the triangle EHB, by adding HGB to each,

the triangles LGB, GEB are equal,

and they are upon the same base GB,
and on the same side of it;

therefore they are between the same parallels,

that is, if L, E were joined, LE would be parallel to GB;
and if a semicircle were described on GB as a diameter,
it would pass through the points E, L;
for the angles at E, L are right angles:

also LE would be a chord parallel to the diameter GB'; therefore the arcs intercepted between the parallels LE, GB are equal,

and consequently the chords EB, LG are also equal; but EB is equal to GM, and GM to GN; wherefore LG, GM, GN, are equal to one another; hence G is the center of the circle inscribed in the triangle BDC. Synthesis. Draw the diagonal BD.

Find G the center of the circle inscribed in the triangle BDC;
through G draw EGN parallel to BC, and FKM parallel to AB.
Then EG and FG bisect the rectangle ABCD.
Draw GL perpendicular to the diagonal BD.
In the triangles GLH, EHB,

the angles GLH, HEB are equal, each being a right angle,.
and the vertical angles LHG, EHB are equal,

also the side LG is equal to the side EB;

therefore the triangle LHG is equal to the triangle EHB. Similarly, it may be proved, that the triangle GLK is equal to the triangle KFD;

therefore the whole triangle KGH is equal to the two triangles ·EHB, KFD`;

and consequently EG, FG bisect the rectangle ABCD.

I.

6. In a given circle, place a straight line equal and parallel to a given straight line not greater than the diameter of the circle.

7. If an equilateral triangle be inscribed in a circle; shew that the radii drawn to its angles trisect the circle.

8. If an equilateral triangle be inscribed in a circle, and a straight line be drawn from the vertical angle to meet the circumference, it will be equal to the sum or difference of the straight lines drawn from the extremities of the base to the point where the line meets the circumference, according as the line does or does not cut the base.

9. If from the angles of an equilateral triangle, perpendiculars be let fall on any diameter of the circumscribing circle, the sum of the perpendiculars on the same side of the diameter will be equal to the perpendicular on the other side.

10. The perpendicular from the vertex on the base of an equilateral triangle, is equal to the side of an equilateral triangle inscribed in a circle whose diameter is the base. Required proof.

11. If an equilateral triangle be inscribed in a circle, and the adjacent arcs cut off by two of its sides be bisected, the line joining the points of bisection shall be trisected by the sides.

12. If an equilateral triangle be inscribed in a circle, any of its sides will cut off one-fourth part of the diameter drawn through the opposite angle.

13. If a triangle be inscribed in one and described about another of each of two concentric circles, shew that it is equilateral.

14. If an equilateral triangle be turned about its center in its own plane, any two positions of the altitude will always make the same angle as those of the sides; and there will be three positions of coincidence of the triangle.

15. About a given triangle, a circle is described, and another triangle is formed by drawing tangents to the circle through the angular points. If this process be continually repeated, the last of the series of triangles so successively formed will be equilateral.

16. If on the sides of any triangle three equilateral triangles be described, and circles inscribed in each of these triangles, the straight lines joining the centers of the circles will form an equilateral triangle.

17. If a circle can be described cutting the sides AB, BC, CA, of a triangle in the points D, E; F, G; H, K; in such a manner that AD=BF= CH, and EB-GC-KA, shew that the triangle is equilateral.

18. ABCP, and A'B'C'P' are two concentric circles, ABC, A'B'C' are any two equilateral triangles inscribed in them. If P, P' be any two points in the circumference of these circles, shew that

A ́P2 + B′P2 + C′P2 = AP12 + BP'2 + CP2.

II.

19. If ABCDEF be an equilateral and equiangular hexagon, and AB, DC be produced to meet in O; BOC is an equilateral triangle.

20. If ABCDEF be a regular hexagon, and AC, BD, CE, DF, EA, FB be joined; another hexagon is formed whose area is onethird of that of the former.

21. BC, CD and DE are contiguous sides of an equilateral hexagon, and BE is joined; prove that the line BE is inclined to one of the sides of the hexagon at an angle which is equal to half a right angle.

22. If an equilateral triangle be inscribed in a circle, and tangents be drawn parallel to the sides; the perimeter of the hexagon thus formed will be two-thirds of the perimeter of the triangle.

23. If two equilateral triangles be described about a circle, they will by their intersections, form a hexagon equilateral, but not generally equiangular.

24. Prove that the area of a regular hexagon is greater than that of an equilateral triangle of the same perimeter.

25. If any six-sided figure be inscribed in a circle, the sum of either three alternate angles is equal to four right angles.

26. If two equilateral triangles be inscribed in a circle so as to have the sides of one parallel to the sides of the other, the figure common to both will be a regular hexagon, whose area and perimeter will be equal to the remainder of the area and perimeter of the two triangles.

27. Determine the distance between the opposite sides of an equilateral and equiangular hexagon inscribed in a circle.

28. Through the angular points of a given triangle, draw straight lines which shall form an equilateral hexagon whose area shall equal twice that of the triangle.

ABC is a triangle, Aa bisects BC in a, and is produced to a', so that aạ' is one-third of Aa, the same construction is made for Bb', Cc. Show that a'b'c is a triangle similar to ABC and equal to it. Shew that the hexagon common to the two triangles is two-thirds of either triangle.

30. If any two consecutive sides of a hexagon inscribed in a circle be respectively parallel to their opposite sides, the remaining sides are parallel to each other.

31. If the alternate sides of a regular hexagon be produced to meet, the figure so formed will be regular, and have an area equal to twice the area of the hexagon.

32. If the alternate sides of a regular hexagon be produced to meet one another, and the angular points of the triangles thus formed be joined, a regular hexagon will be formed, the area of which is equal to three times the area of the original hexagon.

33. If a regular hexagon be inscribed in a circle, six circles equal to it may be described, every one touching the original circle, and two of the others. If the centers of these circles be joined successively, a regular hexagon will be formed, whose area is four times the area of the former: and if the outermost points of the six circles be joined, another hexagon will be formed, whose area is seven times the area of the first.

́34. If ABCDEF be a hexagon such that AB is equal and parallel to DE, and BC to EF, prove that CD is equal and parallel to FA. Also that AD, BE, CF all meet in a point, bisecting each other, and that

AD2 + BE2 + CF2 = AB2 + B C12 + CD2 + AC2 + BD2 + CE”.

35. To inscribe a regular duodecagon in a given circle, and shew that its area is equal to the square on the side of an equilateral triangle inscribed in the circle,