4. In an equilateral triangle the square on the altitude is equivalent to three fourths of the square on the side.

5. In an isosceles triangle, if a perpendicular is drawn from an extremity of the base to the opposite side, then twice the rectangle contained by that side and its segment adjacent to the base is equivalent to the square on the base.

6. In any triangle, if an angle is equal to two thirds of a straight angle, then the square on the side opposite is equivalent to the sum of the squares on the other two sides and the rectangle contained by them.

7. If any point is joined to the four vertices of a rectangle, the sum of the squares on the lines drawn to two opposite vertices is equivalent to the sum of the squares on the other two joining lines.

CONSTRUCTION OF EQUIVALENT POLYGONS

Reduction of polygon to equivalent triangle.

71. PROBLEM 1. To construct a triangle equivalent to a given polygon.

Let ABCDE be the given polygon.

To construct a triangle equivalent to it.

Draw any diagonal AC connecting the ends of two adjacent sides. Through the intermediate vertex B draw BH parallel to this diagonal to

H

meet one of the sides next in order, say EA, in H; and draw CH.

The triangles CAB and CAH are equivalent (33).

To each add the polygon ACDE; then the given polygon ABCDE is equivalent to the polygon EHCD.

The number of sides of the latter polygon is one less than the number of sides of the given polygon.

Repeating this process a set of equivalent polygons having fewer and fewer sides is obtained; and the process ends when a three-sided figure is reached.

Conversion of triangle into equivalent rectangle.

72. PROBLEM 2. To construct a parallelogram equivalent to a given triangle, and having an angle equal to a given angle.

Let ABC be the given triangle, and M the given angle.

To construct a parallelogram equivalent to the triangle ABC, and having an angle equal to M.

Bisect AB at D, and draw DE, making the angle BDE equal to M. Draw CEF parallel to AB, and complete the parallelogram DBFE.

This parallelogram is equivalent to the triangle ABC.
To prove this, draw CD.

The parallelogram DBFE is double the triangle BDC, since they have the same base and equal altitudes (34).

The triangles ADC, DBC are equivalent, having equal bases and the same altitude (29). Hence the triangle ABC is double the triangle DBC.

Therefore the parallelogram DBEE is equivalent to the given triangle ABC (9); and it has an angle equal to the given angle.

Ex. 1. To construct a rectangle that shall be equivalent to a given triangle.

Ex. 2. To construct an isosceles triangle equivalent to a given triangle.

Ex. 3. To construct a parallelogram equivalent to a given parallelogram, and having an angle equal to a given angle.

Conversion of parallelogram into equivalent one.

73. PROBLEM 3. On a given line to construct a parallelogram equivalent to a given parallelogram, and having its angles equal to the angles of this parallelogram.

Let ABCD be the given parallelogram and FK the given line.

品

B

M

N

K

P

It is required to construct on FK a parallelogram equivalent to ABCD and having its angles respectively equal to the angles of ABCD.

Prolong KF, and lay off FE equal to BA, one of the sides of the given parallelogram.

Transfer the figure ABCD to the position EFGH, so that BA may fall on FE (I. 199).

Draw KL parallel to FG to meet HG extended in L. Draw LF, and prolong it to meet HE extended in M. Draw MP parallel to EF, and let it meet the extensions of GF and LK in N and P.

Then NPKF is the required parallelogram.

For it is equivalent to EFGH (36); its angles are equal to the angles of EFGH (I. 126); and it is described on the given line FK.

73 (a). Cor. On a given line to construct a rectangle equivalent to a given rectangle.

Ex. 1. Given one side of a rectangle and the equivalent square, find the adjacent side.

Ex. 2. On a given line to construct a rectangle equivalent to a given triangle (72, 73).

To construct a square equivalent to it.

Prolong AB to E, making BE equal to BC.

Bisect AE at

H (I. 70). With H as center and HE as radius describe the arc EKA. Prolong CB to meet this arc in K.

The square constructed on BK is equivalent to the given rectangle.

To prove this, draw the radius HK.

The rectangle of AB and BE with the square on HB is equivalent to the square on HE (49); that is, to the square on HK, which is equivalent to the sum of the squares on HB and BK (61).

Reject the common square on HB. Then the rectangle of AB and BE is equivalent to the square on BK.

Now the rectangle ABCD is the rectangle of AB and BC, that is, of AB and BE.

Therefore the square on BK is equivalent to the given rectangle.

75. Summary. By combining the constructions given in problems 1, 2, 4, a square can be constructed equivalent to any given polygon.

This square is called its equivalent square; and the process of construction is called squaring the polygon.

Ex. 1.

Ex. 2.

Give the complete construction for squaring a given triangle.
Also for squaring a given quadrangle.

Ex. 3. If the lines AK, KE are drawn, prove that the angle AKE

is equal to the sum of the angles KAE, AEK; and hence that AKE is a right angle.

Ex. 4. Use ex. 3 to construct on a given line a rectangle equivalent to a given square.

[Here AB, BK are given, to find BE. Compare with 73, ex. 1.]

Addition of squares.

76. PROBLEM 5. To construct a square equivalent to the sum of two given squares.

Let KL, MN be the sides of the given squares.

To construct a square equivalent to the sum of the squares on these lines.

Draw CA equal to KL. Erect CB perpendicular to CA and equal to MN. Draw the line AB.

The square on AB is equivalent to the sum of the squares on KL and MN.

For the square on AB is equivalent to the sum of the squares on AC and CB (61) and hence equivalent to the sum of the squares on KL and MN.

Ex. To construct a square equivalent to the sum of three or more given squares.

Subtraction of squares.

77. PROBLEM 6. To construct a square equivalent to the difference of two given squares.

[This is a particular case of I. 136.]

Ex. Show how to construct a square equivalent to the difference of two polygons.