Cor. Problems V. and VI. afford the means of solving the following:

Through a given point, C, without a straight line AB, to draw a second line meeting the first under a given angle M.

LI, which is evidently the line required. Since from any point, I, two lines IL and IL' may be drawn, making with AC an angle equal to M, if M is not a right angle, it follows that the problem admits of two solutions, unless M is a right angle.

PROBLEM VII.

On a given straight line, as a chord, to describe an arc of a circle capable of containing a given angle.

dius EA = EB, describe a circle. The portion ANB is the arc required. For, by construction, BL is a tangent, being perpendicular to the radius EB, and the angle ABL = M, is measured by half the arc AN'B (T. X., C. II.). But each angle in the segment ANB is also measured by half the same arc (T. X.). 5

CONSTRUCTION OF POLYGONS.

PROBLEM VIII.

To construct a triangle when we have given:
I. One side and the two adjacent angles.

II. Two sides and the included angle.

III. The three sides.

It is not necessary to give at length the construction of the first and second cases, since they are so readily deduced from the Corollary to Problem VI. We remark, however, that in the first, in order that the construction may be possible, the sum of the two given angles must be less than two right angles. And in the second case, the construction is always possible, since, after having drawn two lines, forming with each other the required angle, we can always take on these lines portions equal to the given sides.

C

In the third case. Draw AB equal to one of the given sides, and with A as a centre with a radius equal the second of the given sides, describe an arc; also, with B as a centre and with a radius equal to the third side, describe an arc intersecting the former arc in C. Draw CA and CB, and the triangle ABC will evidently satisfy the conditions announced.

A

B

That this construction may be possible, it is necessary that the first side may be less than the sum of the other two sides, and greater than their difference (B. I., T. VII.).

Scholium. If, in the first case, instead of having the two adjacent angles given, we had one of the adjacent angles and the opposite angle, we could, by Prob. V., Scholium II., find the third angle, and then the two adjacent angles would be known.

PROBLEM IX.

To construct a triangle when two sides are given and an angle opposite one of these sides.

First. When the side opposite the given angle is greater

than the adjacent side, the given angle being of any magnitude whatever.

Draw the two lines AB and AC, forming the angle BAC equal to the given angle; take AD equal to the adjacent side, and with D as a centre, and with a radius equal to the oppo

A

D

C

E

B

site side, describe an arc cutting AB in E. Draw DE, and the triangle DAE will fulfil the conditions announced.

Secondly. When the side opposite

the given angle is less than the adjacent side, the given angle being acute.

C

D

A

E' F

E

B

As before, make the angle BAC equal to the given angle, take AD equal to the adjacent side, and with D as a centre, and with a radius equal to the opposite side, describe an arc cutting AB in two points E and E': drawing DE and DE', we have two triangles DAE and DAE' fulfilling the required conditions. If, however, the opposite side is equal to the perpendicular DF, the arc will touch AB, and the two points E and E' will unite with F, and there will be only one triangle, which will be right-angled at F. But if it is less than this perpendicular, the arc cannot either intersect or touch AB, and the construction will be impossible. It is obvious that the construction would be impossible if the given angle were either right or obtuse, so long as the opposite side is less than the adjacent side.

Scholium. It is not necessary to consider the case when the two sides are equal, for the triangle would then be isosceles; and having one angle given, the other angles could readily be found, and then it would be included in Problem VIII.

PROBLEM X.

A regular inscribed polygon being given, to construct a regu lar circumscribed polygon of the same number of sides; and conversely.

First construction. Through the vertices A, B, C, etc., of the inscribed polygon, draw tangents to the circumference, and they will form the polygon required

Second construction. Through the extremities I, K, L, etc., of the radii drawn perpendicular to the sides AB, BC, CD, A etc., of the inscribed polygon, draw tangents which will form the polygon required.

We have already shown that the two polygons thus obtained are equal (T. XVI., S.).

E!

Reciprocally. To obtain the inscribed polygon, when we already know the circumscribed polygon A'B'C'D'E', we join the consecutive points of contact A, B, C, etc., of the circumscribed polygon. These chords, AB, BC, CD, etc., are equal, since their arcs AIB, BKC, CLD, etc., are equal; and they form with each other equal angles, since the arcs which measure them are equal.

Or we could equally as well use the circumscribed polygon A'B'C"D"E", by joining the consecutive points A, B, C, etc., where the straight lines OA", OB", OC", etc., drawn from the centre to the vertices of the circumscribed polygon, meet the circumference.

The polygon ABCDE is regular, since the angles at the centre A"OB", B"OC", etc., being equal, their arcs which measure them are equal, and consequently the chords of these arcs are equal.

PROBLEM XI.

Having given two regular polygons, the one inscribed and the other circumscribed, each of the same number of sides, to inscribe and circumscribe regular polygons of double the number of sides, and of half the number of sides.

Let AB and MN be sides of the

given polygons.

First. To obtain an inscribed polygon of double the number of sides, it is evidently sufficient to join the points A, B, with the point I, the middle of the arc AB, and to repeat the operation for each of the arcs BC, CD, etc.

M

m I n

N

K

L

The chords AI, IB, BL, etc., are equal, since they are subtended by equal arcs.

Secondly. We obtain the corresponding circumscribed polygon, by drawing through the points A, B, two tangents which terminate at the points m, n, on the tangent MN, and repeating the operation at the points C, D, etc.

The line mn is the side of a circumscribed polygon of double the number of sides; and Am, nB are half sides.

In effect, the right-angled triangles OAm, OIm are equal, having the common hypotenuse Om and the side OA=OI; consequently Am=Im, and the angle AOm angle IOm.

We see also that the angle mOn is half the angle at the centre of each of the given polygons, and it is therefore the angle at the centre of the polygon sought.

Thirdly. As to the inscribed and circumscribed polygons of half the number of sides, we may obtain them by drawing through the alternate points A, C, E, etc., chords AC, CE, etc. And through the same points drawing tangents, having no regard to the intermediate points B, D, F, etc.

Scholium. It is evidently necessary in the third case, in order that the problem may be possible, to admit that the given polygons have an even number of sides.

OF CONTACT

PROBLEM XII.

Through a given point without the circumference of a circle

to draw a tangent to this circumference.

Let E be the centre of the given circle, and A the given point. Draw AE, and upon it, as a diameter, describe a circumference which will cut the given circumference in two points M, M', AM, and AM', being drawn, will be the tangents required. For,

E

M

M'