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SUMMARY OF THE RESULTS ARRIVED AT IN

EUCLID, Book II.

The enunciations in the second book of Euclid are so long and confusing that the student will find it better to learn the results of the propositions in the form given below. Prop. 8 is omitted as being unnecessary. The other thirteen propositions may be arranged in the following five groups :(1) One Theorem dealing with two lines AB and X:

Prop. 1. X. AB = X.AC + X.CD + X.DB. (2) Four Theorems dealing with one line AB, divided into two parts

AC, CB :

Prop. 2. ABP = AB.AC + AB.CB.
Prop. 3. AB.AC = AC2 + AC.CB.
Prop. 4. AB2 AC2 + CB2 + 2AC.CB.

Prop. 7. AB2 + CB2 2AB.CB + AC?, (3) Two Theorems, with two cases each, dealing with one line AB divided

equally at P and unequally at Q :

Prop. 5. AQ.QB + PQ2 = PB2 (Q internal).
Prop. 6. AQ.QB + PB2 = PQ2 (Q external).
Prop. 9. AQ? + QB2 = 2(AP2 + PQ2) (Q internal).

Prop. 10. AQ2 + QB2 = 2(AP2 + PQ) (Q external). (4) Two Theorems dealing with triangles :

Prop. 12. AB2 BC2 + CAP + 2BC.CD. (BCA an obtuse angle).

Prop. 13. CAP BC2 + BA2 - 2BC.BD. (ABC an acute angle). (5) Two Problems :

Prop. 11. AB.HB = AH2.

Prop. 14. EH2 = rectilineal figure A. The most important of all these are perhaps Props. 4, 5, 12, 13, and 11.

ADDITIONAL PROPOSITIONS.

1. The square on any straight line is equal to four times the square on half

the line. [Prove by II. 4.] 2. In a right-angled triangle, the square on the perpendicular drawn from

the right angle to the hypotenuse is equal to the rectangle contained by the segments into which the perpendicular divides the hypotenuse.

[I. 47 and II. 4.] 3. If a straight line be divided into two parts, the rectangle contained by the

two parts is a maximum when the parts are equal. (II. 5.] 4. The difference of the squares on two sides of a triangle is equal to twice

the rectangle contained by the base and the line intercepted between the middle point of the base and the foot of the perpendicular drawn from the vertex to the base.

[By I. 47 and II. 5 (see No. 7, p. 139); or by II. 12 and 13.] 5. The sum of the squares on two sides of a triangle is equal to twice the

square on half the base, together with twice the square on the line
joining the vertex to the middle point of the base.
[Draw the perpendicular from the vertex to the base and use II. 12

and 13.] 6. The sum of the squares on the diagonals of a parallelogram is equal to the

sum of the squares on its four sides. [Use Nos. 5 and 1.] 7. The sum of the squares on the four sides of any quadrilateral is equal to

the sum of the squares on its diagonals, together with four times the square on the line joining the middle points of the diagonals.

[Use Nos. 5 and 1.] 8. The square on the straight line drawn from the vertex of an isosceles

triangle to any point in the base is less than the square on either of the
equal sides by the rectangle contained by the segments of the base.
[Draw the perpendicular from the vertex to the base and use 11. 5 and

I. 47 ; or prove by No. 4.] 9. If A, B, C, D be four points taken in order along a straight line, prove that

AB.CD + BC.AD + CA.BD O (Euler's Theorem).

[By II. 1, observing the convention about signs. ] 10. Produce BA to a point K, so that the rectangle AB, KB may be equal to

the square on KA.
[In the figure of Prop. 11 produce AC downwards to L, making EL

equal to EB; and on AL, away from B, describe a square ALMK.] 11. If a straight line be divided internally in medial section, as in Prop. 11,

and if a part be taken from the greater segment equal to the less, prove that the greater segment is divided in medial section. [II. 11 and 11. 3. ] ON THE ALGEBRAICAL SOLUTION OF GEOMETRICAL

PROBLEMS. In the case of Prop. 11 it was shown that the problem could be stated algebraically thus, a (a 2) = x2 ; and in solving this quadratic equation we arrived at the following equation

5a2 X +

2 4 The right-hand member of this equation might have suggested that in order to solve the problem we must find a line whose square is five times the square on half the given line. And this was done, for the square on EF is equal to five times the square on CE, which is equal to half AB. (See p. 145, Ex. 1. (vi).)

Similarly in other cases a geometrical problem may often be stated algebraically, and the solution of an equation may suggest the geometrical solution of the problem. Take for example the following problem :

To divide a given straight line into two parts, so that the square on one part may be equal to twice the square on the other part.

Let a denote the length of the given straight line; and let a denote the length of one part; then a X will be the length of the other part. It is required to find the value of x which satisfies the equation.

2 (a - x) 2

2a 2 4ax + 2x2 .: X2

4ax 2a ? ... 22 – 4ax + (2a) 2

2a ? + 4a 2 :: (2 – 2a) 2 2a

(1) 2a + 2a

(2) Now equation (1) suggests that we must find a line, the square on which is equal to twice the square on the given line ; and equation (2) suggests that the required part x will be found by taking a line equal in length to twice the given line, and then adding or subtracting the length of the line suggested by equation (1).

We know that the square on the diagonal of a square is equal to twice the given square; therefore we get the following solution :A P B

C

Q
Let AB be the given straight line ;
on AB describe the square ADEB, and join DB.

Produce AB to C making AC = 2AB.
Then add to or subtract from AC a part equal to DB,

and we get two segments of AB, viz. AP and AQ
E

such that AP2 = 2PB?, and AQ? = 2BQ?.

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The algebraical solution will not do more than suggest that some line must be found whose length bears a certain ratio to the length of the given line. In the case given, the length of line required is given by the diagonal of the square described on the given line. In other cases other well-known figures will often give the required lengths; but it must depend on the ingenuity of the student to make the right selection.

EXERCISES ON GEOMETRICAL PROBLEMS.

(To be solved algebraically and geometrically.)

1. Construct a rectangle equal to a given rectangle in area, but having one of its sides three times the length of one of the sides of the given rectangle.

2. Construct a rectangle having given its area and perimeter.

3. Divide a given straight line into two parts so that the sum of the squares on them may be equal to a given square.

4. Divide a given straight line AB at C, so that the rectangle contained by BC and a given line may be equal to the square on AC.

5. Divide a given straight line into two parts so that the rectangle contained by the two parts may be equal to the rectangle contained by the sum and difference of the parts.

6. Divide a given straight line both externally and internally into two parts so that the square on one part may be equal to three times the square on the other part.

7. Divide a given straight line both externally and internally so that twice the square on one part may be equal to three times the square on the other part.

8. Divide a straight line into two parts so that the rectangle contained by the two parts may be a maximum.

9. Divide a given straight line both externally and internally into two parts so that the square on one part may be equal to twice the rectangle contained by the whole line and the other part.

MISCELLANEOUS EXERCISES AND RIDERS ON BOOK II.

On Propositions 1–11.

1. If there be two straight lines, each of which is divided into any number of parts, the rectangle contained by the straight lines is equal to the sum of the rectangles contained by each of the parts of the first line and each of the parts of the second line.

2. If D be a point in the hypotenuse of a right-angled triangle, such that the rectangle BD, BC is equal to the square on AC, prove that the rectangle BC, DC is equal to the square on AB.

3. In any right-angled triangle twice the sum of the squares on the three medians is equal to three times the square on the hypotenuse.

4. If the straight line AB be divided into two parts at C so that the square on AB is double the square on AC, and if D be taken in AC so that AD is equal to CB, prove that the square on CD will be double the square on AD.

5. If the straight line AB be divided into three parts at C and D so that AD is equal to CB, and if the square on CB be equal to three times the square on AC, prove that the rectangle AB, CD is equal to twice the square on AC.

6. If a straight line be divided into two pairs of unequal parts, prove that the sum of the squares on the greatest and least of the four parts is greater than the sum of the squares on the other two parts. Express this algebraically.

7. ABC is an equilateral triangle and D any point in BC; prove that the square on BC is equal to the rectangle BD, DC, together with the square on AD.

8. State and prove the theorem corresponding to No. 7 when D is in BC produced.

9. If the straight line AB be divided into any two parts at C, and if AC and CB be bisected at D and E respectively, prove that the square on AE, together with three times the square on EB, is equal to the square on BD, together with three times the square on DA.

10. If the straight line AB be bisected in P and produced to Q so that the rectangle AQ, BQ is equal to twice the rectangle AB, PQ, prove that the square on BQ is equal to the rectangle AB, AQ.

11. Any rectangle is equal to half the rectangle contained by the diagonals of the squares described on two of its adjacent sides.

12. A and B are two fixed points, and P is a point which moves so that the difference of the squares on PA and PB is constant; show that the locus of P is two straight lines at right angles to the line AB,

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