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shown that ADE and PRS are proportional to AEF and PST. Hence,

(1)

ACD ADE: AEF:: PQR: PRS: PST. Now, the sum of the antecedents of (1) is equal to the first polygon, and the sum of the consequents of (1) is equal to the second polygon. Hence, from P. 8, B. 3, we infer that the first polygon is to the second, as any triangle in the first is to the corresponding triangle in the second.

Cor. 2. From the preceding corollary, it follows that two similar polygons are to each other as the squares of any two homologous sides of any two corresponding triangles; but, the homologous sides of the corresponding triangles are the homologous sides or the homologous diagonals of the polygons to which the triangles belong; hence, two similar polygons are to each other as the squares of their homologous sides, or as the squares of their homologous diagonals.

EXERCISES ON BOOK IV.

Let the student demonstrate the following theorems:

1o. The area of a triangle is equal to half the product of its perimeter and the radius of the inscribed circle.

2o. The rectangle of the sum and the difference of two lines is equal to the difference of their squares.

3°. The sum of the squares of two sides of a triangle is equal to twice the square of the third side, increased by twice the square of the line drawn from the middle of the third side to the opposite vertex.

4°. The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

5o. If lines are drawn from any point to the vertices of a rectangle, the sum of the squares of the lines to the extremi

ties of one diagonal is equal to the sum of the squares of the lines to the extremities of the other diagonal.

6o. If a line is drawn from the centre of a circle to any point of any chord, the square of this line, increased by the rectangle of the segments of the chord, is equal to the square of the radius.

APPLICATION OF PRECEDING PRINCIPLES TO THE SOLUTION OF PROBLEMS.

PROBLEM I.

To construct a square equal to the sum of two given squares; also, to construct a square equal to the difference of two given squares.

Let AC and PQ be sides of the given squares, AC being the greater.

1o. Draw two lines KL and KN per

pendicular to each other at K; lay off KL equal to AC and KN equal to PQ; draw LN and in it, as a side, construct the square LO; it will be equal to the sum of the given squares (P. 8).

2o. Draw two lines KL and KT perpendicular to each other at K; lay off KL equal to PQ; from L as a centre, with a radius equal to AC, draw an arc

cutting KT at S, and on KS as a side

M

A

P.

T

K

N

K

construct the square KH; it will be equal to the difference of the given squares (P. 8, Cor. 1).

PROBLEM II.

To construct a triangle equal to a given polygon.
Let ACDEF be the given polygon.

K

A

G

Take AC as a base and prolong it in both directions; draw the diagonals AE and CE. Through F draw FG parallel to EA, and draw EG. Because the triangles AEF and AEG have the common base AE, and because their vertices F and G are on a line FG parallel to that base they are equal (P. 4, Cor. 2). If we take the triangle AEF from the given polygon and then add the equal triangle AEG, the resulting polygon GCDE will be equal to the given polygon. Again draw DK parallel to EC and join the points K and E. Because the triangles ECD and ECK have a common base EC, and because their vertices D and K are in the line DK parallel to that base, the triangles are equal. If we take from the polygon GCDE the triangle CDE, and add the equal triangle CKE, the resulting triangle KEG will be equal to the polygon GCDE, and consequently to the given polygon.

PROBLEM III.

To construct a square equal to a given triangle.

Let GKE be the given triangle, GK being its base and EF its altitude.

Construct a mean proportional CD between GK and FE (Prob. 4, B. 3); then, on CD as a side construct the square AD: it will be the required square; for,

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Scho. By means of Problems 2 and 3 we can construct a square equal to a given polygon, and by the aid of Problem 1 we can construct a square equal to the sum, or to the difference of any two polygons.

PROBLEM IV.

To construct a polygon on a given line as a side, that shall be similar to a given polygon.

Let ACDEF be the given polygon and let PQ be the side which is to be homol

ogous with AC; suppose PQ to be placed parallel to AC.

Draw AD and AE.

D

C

E

R

P

Draw QR parallel to CD, and PR parallel to AD, and let these lines intersect at R; then are the triangles ACD and PQR similar, (P. 15, Cor. 2, B. 3). Draw RS parallel to DE, and PS parallel to AE, and let these lines intersect at S; then are the triangles ADE and PRS similar. In like manner draw ST parallel to EF, and PT parallel to AF, and let these lines intersect at T; then are the triangles AEF and PST similar. From the similarity of the homologous triangles we infer the equality of the corresponding angles of the given polygon and of the polygon PQRST; we also infer the proportionality of their sides taken in the same order; hence, the polygon PQRST is similar to the given polygon.

BOOK V.

PROPERTIES AND MEASURE OF REGULAR POLYGONS AND CIRCLES.

Definition.

83. A regular polygon is a polygon that is both equilateral and equiangular.

PROPOSITION I. THEOREM.

If two regular polygons have the same number of sides they are similar.

For, their corresponding angles are equal, because any angle of either is equal to twice as many right angles as the figure has sides, less four right angles, divided by the number of angles (P. 25, B. 1); furthermore, their corresponding sides are proportional, because all the sides of each polygon are equal to each other; hence, the polygons are similar, which was to be proved.

Cor. If two regular polygons have the same number of sides, their perimeters are proportional to any homologous lines, and their areas are proportional to the squares of these lines.

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