Construct a rectangle N whose base and altitude are EB and AE, then

N=AEX EB,

or

N= M and AE + EB = AB.

PROPOSITION XVII. PROBLEM.

276. To construct a square having a given ratio to a given square.

Let M be the given square, and let the given ratio be that of the lines m and n.

To construct a square which shall have to M the ratio n: m. It is known (from 228) that the perpendicular let fall from any point in the circumference upon the diameter divides it into segments which have the same ratio as the squares of the chords drawn from the same point to the two extremities of the same diameter.

Hence we lay off on a straight line DA =m, and DB = n, and on AB erect (by 183) a semicircumference.

At D erect (by 182) the perpendicular DC, and join CA and CB.

Then (by 228), CA: CB=m: n.

But neither CA nor CB is a side of the square M, that is, we must take a part of CA, say CE, that is equal to a side of M, and find some quantity that has the same ratio to CE that CB has to CA.

It is known (from 213) that a line drawn through E parallel to AB will divide CB into parts having the same ratio as the parts into which E divides CA.

Therefore draw (by 189) EF parallel to AB;

or CF is the side of the square required.

COR. If a side of M is greater than CA, extend CA and CB, and proceed in the same manner.

1. To construct a square equal to the sum of a given triangle and a given parallelogram.

2. To construct an isosceles triangle equivalent to a given triangle, its altitude being given.

PROPOSITION XVIII. PROBLEM.

277. To construct a triangle equivalent to a given polygon. Let ABCDE be the given polygon.

To construct a triangle equivalent

to ABCDE.

If it is possible to construct one E polygon equivalent to another but

with one side less, then a continuation

of this operation would eventually re- G A sult in a triangle.

D

B

C

One side of the polygon, say AB, can be extended without increasing the number of sides, then draw DA, in the effort to

find upon DA and AB produced a triangle equivalent to DEA introducing one line in the place of two, that is DE and EA.

If DA is regarded as the base, then the required triangle must have (by 255) an altitude equal to the distance from E to DA; and since (by 108) parallel lines are everywhere equally distant, the vertex of the required triangle must lie on the parallel to DA drawn through E. Again, if AB produced is to be a side of the triangle, the vertex must also be on AB produced or at G.

Draw DG, and the triangle DGA will be the equivalent of DEA.

Add to DABC the triangle DEA, and we have the original polygon; add to the same figure the equal triangle DGA, and we have the polygon DGBC; therefore DGBC= DEABC, and has one side less.

Draw CF parallel to DB and draw DF; then the triangle DGF will be equivalent to the polygon ABCDE.

EXERCISES.

1. To draw a square equivalent to a given polygon.

2. To construct a square equal to two given polygons.

3. Two similar polygons being given, to construct a similar polygon equal to their sum.

SUGGESTION. Find (by 238) a fourth proportional to EF, AB, and

CD, and it will be the required altitude; then see 196, Ex. 2.

5. When is the last problem impossible?

BOOK V.

REGULAR POLYGONS AND CIRCLES.

278. A Regular Polygon is a polygon which is equilateral and equiangular.

PROPOSITION I. THEOREM.

279. A circle may be circumscribed about, or inscribed within, any regular polygon.

Let ABCDE be a regular polygon.

1. To prove that a circle may be circumscribed about it. Let A, B, and C be any three vertices, and through them. pass (by 194) a circle; let its centre be at O. Join OA, OB, OC, OD, and OE.

Since the polygon is equiangular,

ZABC/BCD,

and since OB = OC, OBC = ≤ OCB.

or

Subtracting these equal angles,

ZABC-ZOBC= ZBCD- OCB,

ZOBA=2OCD;

=

OB being

OA.

therefore the triangles OCD and ABO have OC = radii, AB = CD sides of the regular polygon, and ZOBA = OCD. They are therefore equal, and OD Hence the circle passing through A, B, and C, also passes through D.

=

In the same manner it can be shown to pass through E.

2. To prove that a circle may be inscribed in ABCDE. Since AB, BC, CD, DE, and EA are equal chords, they are (by 148) equally distant from the centre O.

Hence if a circle be described with O as a centre, and a radius equal to the perpendicular distance from 0 to one of the sides, the circumference will touch all the sides of the polygon.

Q.E.D.

280. The Centre of a regular polygon is the common centre O of the circumscribed and inscribed circles.

281. The Radius of a regular polygon is the radius OA of the circumscribed circle.

282. The Apothem of a regular polygon is the radius OF of the inscribed circle.

283. The Angle at the centre is the angle included by the radii drawn to the extremities of any side.

284. COR. 1. Each angle at the centre of a regular polygon is equal to four right angles divided by the number of sides of the polygon.

Since the triangles OAB, OBC, OCD, etc., are equal, the angles AOB, BOC, COD, etc., are equal.

Therefore each angle is equal to four right angles (by 49) divided by the number of sides.

285. COR. 2. If a regular inscribed polygon is given, the tangents at the vertices of the given polygon form a regular circumscribed polygon of the same number of sides.