COR. Conversely to Theorems 9, 10, 11, the angle opposite a side of a triangle is a right angle, an obtuse angle, or an acute angle, according as the square on that side is equal to, greater than, or less than the sum of the squares on the other two sides.

For in this group of Theorems the hypotheses are exhaustive and the conclusions are mutually exclusive. Hence the Rule of Conversion is applicable. Introd. § 9.

Ex. 22.

Shew that Theor. 7 is a limiting case of Theor. 11.

THEOR. 12. The sum of the squares on two sides of a triangle is double the sum of the squares on half the base and on the line joining the vertex to the middle point of the base,

Let ABC be a triangle, D the middle point of the base BC: then shall the sum of the squares on AB and AC be double the sum of the squares on BD and DA.

If AD is at right angles to BC,

B

D

the square on AB is equal to the sum of the squares on BD and DA,

II. 9. and the square on AC is equal to the sum of the squares on CD

II. 9.

and DA,

that is, to the sum of the squares on BD and DA, since CD is equal to BD;

therefore the sum of the squares on AB and AC is double the sum of the squares on BD and DA.

But if AD is not at right angles to BC,

let ADB be an obtuse angle, and let DE be the projection on BC of AD.

Because ADB is an obtuse angle,

II. 10.

therefore the square on AB is greater than the sum of the squares on BD and DA by twice the rectangle BD, DE; and because ADC is an acute angle,

therefore the square on AC is less than the sum of the squares on CD and DA by twice the rectangle CD, DE, 11. II. that is, the square on AC is less than the sum of the squares on BD and DA by twice the rectangle BD, DE,

since CD is equal to BD;

therefore the sum of the squares on AB and AC is double the sum of the squares on BD and DA.

Q.E.D.

Ex. 23. The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals.

Ex. 24. The sum of the squares on the sides of any quadrilateral

is greater than the sum of the squares on the diagonals by four times the square on the line joining the middle points of the diagonals.

THEOR. 13. If a straight line is divided internally or externally at any point, the sum of the squares on the segments is double the sum of the squares on half the line and on the line between the point of division and the middle point of the line.

Let the straight line AB be bisected at C and divided internally or externally at D:

E

B

then shall the sum of the squares on AD and DB be double the sum of the squares on AC and CD.

EC.

First Proof. Draw DE at right angles to AB, join EA, EB,

The sum of the squares on AE and EB is double the sum of the squares on AC and CE;

II. 12.

but the square on AE is equal to the sum of the squares on AD and DE,

II. 9.

the square on EB is equal to the sum of the squares on BD and DE,

II. 9.

and the square on CE is equal to the sum of the squares on CD and DE;

II. 9.

therefore the sum of the squares on AD and DB together with twice the square on DE is equal to double the sum of the squares on AC, CD and DE,

therefore the sum of the squares on AD and DB is double the sum of the squares on AC and CD.

Q.E.D.

Second Proof. Because AD is the sum of AC and CD, therefore the square on AD is greater than the sum of the squares on AC and CD by twice the rectangle AC, CD; and because BD is the difference of BC and CD,

II. 6.

that is, of AC and CD, since AC is equal to BC, therefore the square on BD is less than the sum of the squares on AC and CD by twice the rectangle AC, CD; II. 7. therefore the sum of the squares on AD and BD is double the sum of the squares on AC and CD.

Q.E.D.

COR. The sum of the squares on the sum and on the difference of two given lines is double the sum of the squares on those lines.

Ex. 25. The difference of the squares on the sum and difference of two given lines is equal to four times the rectangle contained by the lines.

Ex. 26. Shew that Theor. 13 is a limiting case of Theor. 12.

SECTION II.

PROELEMS.

PROB. 1. To construct a parallelogram equal to a given triangle and having one of its angles equal to a given angle.

Let ABC be the given triangle, and D the given angle:

it is required to construct a parallelogram equal to ABC, and having an angle equal to D.

Bisect BC at E,

E

I. Prob. 4.

at the point E in CE make the angle CEF equal to D, I. Prob. 5.

Then each of the triangles AEB, AEC, is equal to the triangle FEC,

II. 2, Cor. I.

therefore the triangle ABC is double the triangle FEC; and because the diagonal FC of the parallelogram FECG bisects

it,

I. 29.

therefore the parallelogram FECG is also double the triangle FEC;