the side D E produced both ways for the base; through C draw CH parallel to B D, and join B H; through B draw B K parallel to A H, and join A K; then A K will be one side of the triangle. Proceeding in like manner to eliminate the angular points F and G, the other side, A L, of the triangle will be determined.

Reason. The triangle B H D is (Theor. 29) made equal to the triangle B C D, and, therefore, the figure A BHEF G is equivalent to the figure A B CDEFG. Again, the triangle A K H is made equal to the triangle A B H, and A K E F G is equivalent to the figure A BHEF G, and thus the sides A B, B C, C D have been replaced by the single side A K. In like manner, the sides E F, F G, G A are replaced by the single side A L, and the seven-sided figure A B C D E F G is reduced to the triangle A L K.

From the various problems which have already been solved on the construction of equivalent triangles, there will be no difficulty in modifying this problem so as to place the vertex at a point in one of the sides, or in any other desired position.

If the base L K be bisected in the point M, it is manifest that the line A M, which bisects the triangle A L K, will also bisect the figure A B C D E F G. Also the figure may be bisected by a line drawn from a given point in one of the sides, by constructing an equivalent triangle, with the given point for its vertex, and bisecting the triangle by a line drawn through this vertex. And, again, the figure may be bisected by lines drawn from any given point within it, by constructing an equivalent triangle, with its vertex at one of the angular points, or on one of the sides of the figure, and having the given point within it, since the lines drawn through the given point to bisect this triangle will also bisect the figure.

235. PROBLEM 120.-To construct a square equivalent to a given rectilineal figure.

Reduce the figure to a triangle, and construct a square equivalent to the triangle.

EXERCISES.

(1.) On a line 2 inches long, construct a regular hexagon; reduce it to a triangle; construct (1) an equivalent equilateral triangle; (2) an equivalent square.

(2.) Inscribe a regular octagon in a square whose side is 8 inches; construct a square equal to the octagon.

(3.) Construct a polygon, A B C DEF, from the following

Bisect the polygon (1) by a line drawn through A; (2) by lines drawn from P, the vertex of the equilateral triangle described upon A B. State the lengths of the bisecting lines; also the length of the side A F; the length of the diagonal A D; and the magnitude of the angle E FA.

236. PROBLEM 121.-Upon a given straight line to construct a polygon similar and similarly situated to a given polygon.

METHOD I.-C D, Fig. 187, being the side of the given polygon corresponding to the side of the required figure, which

the given line A B is to form, divide the polygon into triangles by lines drawn from one extremity D of C D to the other angular points of the polygon. Make the angles BAF and A BF respectively equal to the angles D CE and CDE; then the third angle A F B will be equal to the third angle C E D, and the triangle A F B will be equiangular, and therefore similar (Theor. 18) to the triangle CED. Again, make the angles BFH and FB H respectively equal to D E G and EDG; then the figure A FHB will be similar and similarly situated to the figure CEG D.

Reason. The triangles A F B and F H B being respectively similar and similarly situated to the triangles CED and E GD, the angle AF B is equal to the angle C E D, and B F H to DEG; therefore the whole angle A F H is equal to the whole angle CE G. In like manner, the angles A B H and C D G are equal. Also the angles at A and H are respectively equal to the angles at C and G; and, therefore, the figures AFHB and CEGD are equiangular. Again, since A Fis to F B as CE to E D, and F B to FH as ED to E G; therefore A F is to F H as CE to E G, and the sides about the equal angles A FH and CE G, are proportional. In like manner, the sides about the equal angles ABH and CDG are proportional; also from the similarity of the triangles, the sides about the angles at A and H are respectively proportional to the sides about the angles at C and G. Hence the figures A FH B and C E GB, being equiangular, and having their sides about the equal proportionals, are (24) similar.

Proceeding in the same manner to add to the figure A FH B other triangles, similar and similarly situated to the remaining triangle into which the given polygon CEG KD has been divided, a polygon will be described on A B, similar and similarly situated to the given polygon.

METHOD II.-In this, as in Method I, the required polygon is formed by building it up of triangles similar and similarly situated to those into which the given polygon is divided as before; but instead of forming these triangles, by making them equiangular, the sides about each of their angles are made proportional. Upon a straight line, Q R, Fig. 188, take Q R equal to C D, and Q S equal to AB; from centre Q, with radii QR and Q S, describe concentric arcs, R U SW; in R U place the chords RT and R U, equal respectively to the sides C E and DE of the triangle CED; join QR, QT, cutting the arc SW in V and W; then the chords S V and S W will be the required lengths for the sides A F, B F, of the triangle A F B.

and

Reason.-Q R being equal to Q T, and QS to Q V, the sides of the triangles Q S V and Q R T about the common angle at Q

are proportionals; therefore (Eu. VI. 6), these triangles are similar, and QR is to R T as QS to VS. In like manner, QR is to RU as Q S to S W. Hence, QR, R T, and R U, being equal respectively to C D, C E, and D) E, and Q S being equal to A B, if A F and B F be made respectively equal to S V and S W, the sides of the triangles A F B and C ED, about each of their angles, will be proportionals; and, therefore (Eu. VI. 5), these triangles will be equiangular and similar. In like manner, by setting off on the arc R U chords equal to the sides of the other triangles into which the given polygon is divided, chords upon the arc S W will be determined equal to the required lengths of the sides of the triangles, which will build up the similar and similarly situated figure on A B.

EXERCISES.

(1.) The sides of a triangle being 2 inches, 3 inches, and 5 inches long respectively, describe a similar triangle, whose longest side shall be the side of an equilateral triangle described in a circle of 2 inches diameter.

(2.) Construct a polygon, A B C D E F G, from the following conditions:

On a line, 1.4 inches long, construct a polygon similar to the above polygon, so that the given line may form the side corresponding to D E. Give the measurements of D E and CF, and of the corresponding lines on the similar figure described on the given line.

(3.) The sides A B, A C, of a triangle are each 2 inches long, and contain an angle of 30°. A square, A DE C, is described on the side A C, and another square, B C F G, on the base B C; and E F being joined, an irregular hexagon, A DEFG B, is formed. Take a line 2 inches long to form the side corresponding to E F, and construct upon it a similar hexagon. Measure the length of the diagonal D G, and of the corresponding diagonal in the similar hexagon.-4ns. 4.2 inches; 3.5 inches.

237. PROBLEM 122.-To construct a rectilineal figure which shall be similar to one, and equivalent to another given rectilineal figure.

Reduce (Prob. 119) half the figure F, Fig. 189, to which the required figure is to be similar, to a triangle having its base in a line with one of its sides A B, and construct (Prob. 109) on A B an equivalent right-angled triangle A B C. Again, reduce half the figure G, to which the required figure is to be equal, to a triangle, and construct an equivalent right-angled triangle, CB D, on B C; then will A B D be a straight line, and the triangles A B C, C B D, will have the same altitude as B C. Find B E, the mean proportional between A B and B D; on BE, or on any line equal to B E, construct a figure similar

and similarly situated to the given figure, so that B E may correspond to A B; then the figure thus described will be the figure required.

Reason.-Because A B is to B E as B E to BD; therefore (Theor. 32), as A B is to B D, so is the figure on A B to the similar and similarly described figure on B E; but as A B is to B D, so also is the triangle A B C to the triangle CBD (Theor. 29); therefore, as the figure on A B is to the similar figure on B E, so is the triangle A B C to the triangle C B D. But the trianglo A B C was made equivalent to the figure on A B; therefore,