PROBLEM XXXI.

335. To construct a rectangle equivalent to a given square, and having the sum of its adjacent sides equal to a given line.

Let the straight line AB be equal to the sum of the adjacent sides of the required rectangle.

Upon A B as a diameter describe

D

a semicircle; at the point A, draw A

E B

A D perpendicular to A B, making A D equal to the side of the given square; then draw the line DC parallel to the diameter A B. From the point C, where the parallel meets the circumference, draw CE perpendicular to the diameter; A E and E B will be the sides of the rectangle required.

For their sum is equal to AB; and their rectangle AEX EB is equivalent to the square of CE, or to the square of AD (Prop. XXXII. Cor., Bk. IV.); hence, this rectangle is equivalent to the given square.

336. Scholium. The problem is impossible, when the distance A D is greater than the half the given line A B, for then the line D C will not meet the circumference.

PROBLEM XXXII.

337. To construct a rectangle that shall be equivalent to a given square, and the difference of whose adjacent sides shall be equal to a given line.

Let the straight line AB be equal to the difference of the adjacent sides of the required rectangle.

Upon A B as a diameter, describe a circle. At the extremity of the diameter, draw the tangent AD, making it equal to the side of the given square.

Through the point D and the centre C draw the secant DCF, intersecting the circumference in E; then DE and D F will be the adjacent sides of the rectangle required.

For the difference of these lines is equal to the diameter EF or AB; and the rectangle DE X DF is equal

D

E

F

B

to A D (Prop. XXXV. Cor., Bk. IV.); hence it is equivalent to the given square.

PROBLEM XXXIII.

338. To construct a square that shall be to a given square as one given line is to another given line.

Draw the indefinite line AB,

on which take AC equal to one
of the given lines, and CB equal A
to the other. Upon A B as a di-
ameter, describe a semicircle, and

E

D

F

at the point C draw the perpendicular C D, meeting the circumference in D. Through the points A and B draw the straight lines D E, D F, making the former equal to the side of the given square; and through the point E draw EF parallel to AB; DF will be the side of the square required.

For, since E F is parallel to A B,

DE: DF:: DA:DB;

consequently (Prop. XV. Bk. II.),

DE2: DF:: DA: DB2.

But in the right-angled triangle ADB the square of A D is to the square of D B as the segment AC is to the segment CB (Prop. XI. Cor. 3, Bk. IV.); hence,

DE2: DF2:: AC: CB.

But, by construction, D E is equal to the side of the given square; also, A C is equal to one of the given lines, and CB to the other; hence, the given square is to that constructed on D F as the one given line is to the other.

PROBLEM XXXIV.

339. Upon a given base to construct an isosceles triangle, having each of the angles at the base double the vertical angle.

Let A B be the given base.

Produce A B to some point C till the rectangle ACX BC shall be equivalent to the square of AB (Prob. XXXII.); then, with the base A B and sides each equal to A C, construct the isosceles triangle DA B, and the angle A will double the angle D.

For, make DE equal to A B, or make A E equal to BC, and join E B. Then, by construction,

AD: AB::AB: AE;

for A E is equal to BC; consequently the triangles DAB, BAE have a common angle, A, contained by proportional sides; hence they are similar (Prop. XXIV. Bk. IV.); therefore these triangles are both isosceles, for D A B is isosceles by construction, so that A B is equal to EB; but A B is equal to DE; consequently D E is equal to E B, and therefore the angle D is equal to the angle EBD; hence the exterior angle A E B is equal to double the angle D, but the angle A is equal to the angle A EB; therefore the angle A is double the angle D.

PROBLEM XXXV.

340. Upon a given straight line to construct a polygon similar to a given polygon.

the straight line F G, make the angle GFH equal to the angle BAC; and at the point G make the angle F G II equal to the angle A B C. The lines FH, GH will cut each other in H, and FGH will be a triangle similar to ABC. In the same manner, upon FH, homologous to A C, construct the triangle FIH similar to A DC; and upon FI, homologous to AD, construct the triangle FIK similar to A D E. The polygon FGHIK will be similar to ABCDE, as required.

For these two polygons are composed of the same number of triangles, similar each to each, and similarly situated (Prop. XXX. Cor., Bk. IV.).

PROBLEM XXXVI.

341. Two similar polygons being given, to construct a similar polygon, which shall be equivalent to their sum or their difference.

Let A and B be

two homologous sides of the given polygons.

Find a square equal to the sum or to the difference of the

squares described up

A

B

on A and B; let x be the side of that square; then will x in the polygon required be the side which is homologous to the sides A and B in the given polygons. The polygon itself may then be constructed on x, by the last problem.

For similar figures are to each other as the squares of their homologous sides; but the square of the side x is equal to the sum or the difference of the squares described upon the homologous sides A and B; therefore the figure described upon the side x is equivalent to the sum or to the difference of the similar figures described upon the sides A and B.

PROBLEM XXXVII.

342. To construct a polygon similar to a given polygon, and which shall have to it a given ratio.

Let A be a side of the given polygon. Find the side B of a square, which is to the square on A in the given ratio of the polygons (Prob. XXXIII.).

Upon B construct a polygon similar to the given polygon (Prob. XXXV.), and B will be the polygon required.

A

For the similar polygons constructed upon A and B have the same ratio to each other as the squares constructed upon A and B (Prop. XXXI. Bk. IV.).

PROBLEM XXXVIII.

343. To construct a polygon similar to a given polygon, P, and which shall be equivalent to another polygon, Q.

Find M, the side of a square, equivalent to the polygon P, and N, the side of a square equivalent to the polygon Q. Let x be a fourth proportional to the three given lines

A

Р

B

M, N, A B; upon the side x, homologous to A B, describe a polygon similar to the polygon P (Prob. XXXV.); it will also be equivalent to the polygon Q.