PROP. XLV.-PROBLEM.

To construct a parallelogram equal to a given rectilineal figure (ABCD), and having an angle equal to a given rectilineal angle (X).

Sol. Join BD. Construct a parallelogram EG [XLII.] equal to the triangle ABD, and having the angle E equal to the given angle X; and to the right

line GH apply the parallelogram HI equal to the triangle BCD, and having the angle GHK equal to X [XLIV.], and so on for additional triangles if there be any. Then EI is a parallelogram fulfilling the required conditions.

Dem.-Because the angles GHK, FEH are each equal to X (const.), they are equal to one another: to each add the angle GHE, and we have the sum of the angles GHK, GHE equal to the sum of the angles FEH, GHE; but since HG is parallel to EF, and EH intersects them, the sum of FEH, GHE is two right angles [XXIX.]. Hence the sum of GHK, GHE is two right angles; therefore EH, HK are in the same right line [XIV.].

Again, because GH intersects the parallels FG, EK, the alternate angles FGH, GHK are equal [xxIx.]: to each add the angle HGI, and we have the sum of the angles FGH, HGI equal to the sum of the angles GHK, HGI; but since GI is parallel to HK, and GH intersects them, the sum of the angles GHK, HGI is equal to two right angles [XXIX.]. Hence the sum of the

angles FGH, HGI is two right angles; therefore FG and GI are in the same right line [xiv.].

Again, because EG and HI are parallelograms, EF and KI are each parallel to GH; hence [xxx.] EF is parallel to KI, and the opposite sides EK and FI are parallel; therefore EI is a parallelogram; and because the parallelogram EG (const.) is equal to the triangle ABD, and HI to the triangle BCD, the whole parallelogram EI is equal to the rectilineal figure ABCD, and it has the angle E equal to the given angle X. Hence EI is a parallelogram fulfilling the required conditions.

It would simplify Problems XLIV., XLV., if they were stated as the constructing of rectangles, and in this special form they would be better understood by the student, since rectangles are the simplest areas to which others are referred.

Exercises.

1. Construct a rectangle equal to the sum of two or any number of rectilineal figures.

2. Construct a rectangle equal to the difference of two given figures.

PROP. XLVI.-PROBLEM.

On a given right line (AB) to describe a square.

Sol.-Erect AD at right angles to AB [x1.], and make it equal to AB [.]. Through D

D draw DC parallel to AB [xxx1.], and through B draw BC parallel to AD; then AC is the square required.

Dem.-Because AC is a parallelogram, AB is equal to CD [xXXIV.]; but AB is equal to AD (const.); therefore AD is equal to CD, and AD is A

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equal to BC [XXXIV.]. Hence the four sides are equal; therefore AC is a lozenge, and the angle A is a right angle. Therefore AC is a square (Def. xxx.).

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Exercises.

1. The squares on equal lines are equal; and, conversely, the sides of equal squares are equal.

2. The parallelograms about the diagonal of a square are squares.

3. If on the four sides of a square, or on the sides produced, points be taken equidistant from the four angles, they will be the angular points of another square, and similarly for a regular pentagon, hexagon, &c.

4. Divide a given square into five equal parts; namely, four right-angled triangles, and a square.

PROP. XLVII.-THEOREM.

In a right-angled triangle (ABC) the square on the hypotenuse (AB) is equal to the sum of the squares on the other two sides (AC, BC).

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Dem. On the sides AB, BC, CA describe squares [XLVI.]. Draw CL parallel to AG. Join CG, BK. Then because the angle ACB is right (hyp.), and ACH is right, being the angle of a square, the sum of the angles ACB, ACH is two right angles; therefore BC, CH are in the same right line [XIV.]. In like manner AC, CD are in the same right line. Again, because BAG is the

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angle of a square it is a right angle: in like manner CAK is a right angle. Hence BAG is equal to CAK: to each add BAC, and we get the angle CAG equal to

KAB. Again, since BG and CK are squares, BA is equal to AG, and CA to AK. Hence the two triangles CAG, KAB have the sides CA, AG in one respectively equal to the sides KA, AB in the other, and the contained angles CAG, KAB also equal. Therefore [IV.] the triangles are equal; but the parallelogram AL is double of the triangle CAG [XLI.], because they are on the same base AG, and between the same parallels AG and CL. In like manner the parallelogram AH is double of the triangle KAB, because they are on the same base AK, and between the same parallels AK and BH; and since doubles of equal things are equal (Axiom vi.), the parallelogram AL is equal to AH. In like manner it can be proved that the parallelogram BL is equal to BD. Hence the whole square AF is equal to the sum of the two squares AH and BD.

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Or thus: Let all the squares be made in reversed directions. Join CG, BK, and through C draw OL parallel to AG. Now, taking the BAC from the right LS BAG, CAK, the remaining s CAG, BAK are equal. Hence the As CAG, BAK have the side CA AK, and AG AB, and the Z CAG= BAK; therefore [Iv.] they are equal; and since [XLI.] the Os AL, AH are respectively the doubles of these triangles, they are equal. In like manner thes BL, BD are equal; hence the whole square AF is equal to the sum of the two squares AĤ, BD.

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This proof is shorter than the usual one, since it is not necessary to prove that AC, CD are in one right line. In a similar way the Proposition may be proved by taking any of the eight figures formed by turning the squares in all possible directions. Another simplification of the proof would be got by considering that the point A is such that one of the As CAG, BAK can be turned round it in its own plane until it coincides with the other; and hence that they are congruent.

Exercises.

1. The square on AC is equal to the rectangle AB . AO, and the square on BC = AB. BO.

2. The square on CO = AO. OB.

3. AC2 BC2=AO2 - BO2.

4. Find a line whose square shall be equal to the sum of two given squares.

5. Given the base of a triangle and the difference of the squares of its sides, the locus of its vertex is a right line perpendicular to the base.

6. The transverse lines BK, CG are perpendicular to each other.

7. If EG be joined, its square is equal to AC2 + 4BC2.

8. The square described on the sum of the sides of a rightangled triangle exceeds the square on the hypotenuse by four times the area of the triangle (see fig., XLVI., Ex. 3). More generally, if the vertical angle of a triangle be equal to the angle of a regular polygon of n sides, then the regular polygon of n sides, described on a line equal to the sum of its sides, exceeds the area of the regular polygon of n sides described on the base by n times the area of the triangle.

9. If AC and BK intersect in P, and through P a line be drawn parallel to BC, meeting AB in Q; then CP is equal to PQ.

10. Each of the triangles AGK and BEF, formed by joining adjacent corners of the squares, is equal to the right-angled triangle ABC.

11. Find a line whose square shall be equal to the difference of the squares on two lines.

12. The square on the difference of the sides AC, CB is less than the square on the hypotenuse by four times the area of the triangle.

13. If AE be joined, the lines AE, BK, CL, are concurrent.

14. In an equilateral triangle, three times the square on any side is equal to four times the square on the perpendicular to it from the opposite vertex.

15. On BE, a part of the side BC of a square ABCD, is described the square BEFG, having its side BG in the continuation of AB; it is required to divide the figure AGFECD into three parts which will form a square.

16. Four times the sum of the squares on the medians which bisect the sides of a right-angled triangle is equal to five times the square on the hypotenuse.