GIVEN—a, the side of the given square R, and AB, the given line.

TO CONSTRUCT—a rectangle equivalent to R and having its base and altitude together equal to AB.

Upon AB as a diameter construct a semicircle.

Draw CD parallel to AB and at a distance from it equal

to a.

From D the intersection of CD with the circumference draw DX perpendicular to AB.

The rectangle having AX for its altitude and XB for its base is the required rectangle.

417. Remark.-416 may be stated: To find two straight lines of which the sum and product are given.

418. CONSTRUCTION. To construct a rectangle equivalent to a given square, and having the difference of its base and altitude equal to a given line.

GIVEN

a, the side of the square R, and the line AB.

TO CONSTRUCT—a rectangle equivalent to R, and having the difference of its base and altitude equal to AB.

Upon AB as a diameter construct a circumference.
At A draw the tangent AC equal to a.

Draw CXY through the centre meeting the circumference

in X and Y.

Then the rectangle having its base equal to CY and its altitude equal to CX is the required rectangle.

Also XY, the difference between CY and CX, is a diameter of the circle, and therefore equal to AB.

Q. E. F.

419. Remark.-§ 418 may be stated: To find two straight lines of which the difference and product are given.

420. CONSTRUCTION. To construct a polygon similar to a given polygon and equivalent to another given polygon.*

* Pythagoras (about 550 B.C.) first solved this problem.

TO CONSTRUCT-
-—a polygon similar to P and equivalent tó Q.

Construct squares equivalent to P and Q.

Let n and m be the sides of these squares.

8415

From any point O draw two lines OM and ON, and on these lay off OC equal to m and OD equal to n. On OD lay off OS equal to a, a side of P.

Draw parallels giving the fourth proportional OT. § 282 Upon OT, or x, as a side homologous to a, construct a polygon X similar to P. It will also be equivalent to Q. m2 sq. on m

Proof:

2

X x2

-=

P α

2

n sq. on n P

(Why?)

Therefore X is equivalent to Q and is similar to P by construction.

Q. E. F.

PROBLEMS OF DEMONSTRATION

421. The square on the base of an isosceles triangle, whose vertical angle is a right angle, is equivalent to four times the triangle.

422. A quadrilateral is divided into two equivalent triangles by one of its diagonals, if the other diagonal is bisected by the first.

423. The four triangles formed by drawing the diagonals of a parallelogram are all equivalent.

424. If from the middle point of one of the diagonals of a quadrilateral straight lines are drawn to the opposite vertices, these two lines divide the figure into two equivalent parts.

425. If the sides of any quadrilateral are bisected and the points of bisection successively joined, the included figure will be a parallelogram equal in area to half the original figure.

426. A trapezoid is divided into two equivalent parts by the straight line joining the middle points of its parallel sides.

427. The triangle formed by joining the middle point of one of the non-parallel sides of a trapezoid to the extremities of the opposite side is equivalent to one-half the trapezoid.

428. If the three sides of a right triangle are the homologous sides of similar polygons described upon them, then the polygon described upon the hypotenuse is equivalent to the sum of the polygons described upon the other two sides.

429. If M is the intersection of the medians of a triangle ABC, the triangle AMB is one-third of ABC.

430. If from the middle point of the base of a triangle lines parallel to the sides are drawn, the parallelogram thus formed is equivalent to one-half the triangle.

431. Any straight line drawn through the intersection of the diagonals of a parallelogram divides the parallelogram into two equivalent parts.

432. The square described upon the sum of two straight lines is equivalent to the sum of the squares described upon the two lines plus twice their rectangle.

Hint.-Let AB and BC be the given lines.

433. The square described upon the difference of two straight lines is equivalent to the sum of the squares described upon the two lines minus twice their rectangle.

Hint.-Let AB and BC be the given lines.

434. The rectangle whose sides are the sum and the difference of two straight lines is equivalent to the difference of the squares described upon the two lines.

B

Hint.-Let AB and BC be the given lines.

Question.-To what three formulas of algebra* do the last three problems correspond?

* Euclid gave the geometric proofs of S$ 432-4; but though he may have translated them into algebra, he was probably not acquainted with the algebraic proof. To-day we find it easier to obtain the algebraic formulas first, and then give them the geometric interpretation. This is true in a multitude of cases where the opposite was true among the Greeks.