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(Const.) Describe (I. 45) the rectangular parallelogram BCDE equal to the rectilineal figure A. If, then, the sides of it BE, ED are equal to one another, it is a square, and what was required is done; but if they are not equal, produce one of them BE to F, and make EF equal to ED, and bisect BF in G; and from the centre G, at the distance GB or GF, describe the semicircle BHF, and produce DE to H, and join GH. (Dem.) Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the rect..

EF angle BE · EF, together with the square on EG, is equal to the square on GF (II. 5). But GF is equal to GH; therefore the rectangle BE · EF, together with the square on EG, is equal to the square on GH. But the squares on HE and EG are equal to the square on GH (I. 47); therefore the rectangle BEEF, together with the square on EG, is equal to the squares on HE and EG. Take away the square on EG, which is common to both, and the remaining rectangle BE · EF is equal to the square on EH; but the rectangle contained by BE. EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square on EH. But BD is equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square on EH; wherefore the side of a square has been found equal to the given rectilineal figure A, and the square described upon EH is therefore equal to A.

PROPOSITION A. THEOREM. If one side of a triangle be bisected, the sum of the squares on the other two sides is double of the square on half the side bisected, and of the square on the line drawn from the point of bisection to the opposite angle of the triangle.

Given a triangle ABC, of which the side BC is bisected in D, and DA drawn to the opposite angle; to prove that the squares on BA and AC are together double of the squares on BD and DA.

(Const.) From A draw AE perpendicular to BC. (Dem.) Because BEA is a right angle, the square on AB is equal to the squares on BE and EA (I. 47); for the same reason, the square on AĆ is equal to the two squares on CE and EA; therefore the squares on BA and AC are equal to the squares on BE and EC, together with twice the square on EA. But because

the line BC is cut equally in D, and unequally in E, the squares on BE and EC are equal to twice the squares on BD and DE (II. 9); therefore the squares on BA and AC are equal to twice the square on BD, together with twice the squares on DE and EA. Now, the squares on DE and EA are equal to the square on DA (I. 47), and therefore twice the squares on DE and EA is equal to twice the square on DA; wherefore, also, the squares on BA and AC are equal to twice the square on BD, together with twice the square on DA.

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PROPOSITION B. THEOREM. The sum of the squares on the diameters of any parallelogram is equal to the sum of the squares on the sides of the parallelogram.

Given a parallelogram ABCD, of which the diameters are AC and BD; to prove that the sum of the squares on AC and BD is equal to the sum of the squares on AB, BC, CD, and DA.

(Dem.) Let AC and BD intersect one another in E; because the vertical angles AED, CEB are equal (I. 15),

and also the alternate angles EAD, ECB (I. 29), the triangles ADE, CEB, have two angles in the one equal to two angles in the other, each to each; but the sides AD and BC, which are opposite to equal angles in these triangles, are Ś also equal (I. 34); therefore the other sides which are opposite to the equal angles are equal (I. 26);

namely, AE to EC, and ED to EB. Since, therefore, BD is bisected in E, the squares on BA and AD are equal to twice the square on BE, together with twice the square on EA (II. A); and for the same reason the squares on BC and CD are equal to twice the square on BE, together with twice the square on EC; that is, on EA, because EC is equal to EA; therefore the four squares on BA, AD, DC, and CB are equal to four times the squares on BE and EA. But the square on BD is equal to four times the square on BE, because BD is double of BE (II. 8, Cor. 2); and for the same reason, the square on AC is equal to four times the square on AE; wherefore, also, the squares on BD and AC are equal to the four squares on BA, AD, DC, and CB. COR.-From the demonstration, it is manifest that the diameters

of every parallelogram bisect one another.

PROPOSITION C. THEOREM. The difference between the squares on the two sides of a triangle, is equal to the difference between the squares on the segments of the base, formed by a perpendicular upon it from the vertex.

- Given a triangle ABC, and CD a perpendicular on the base AB; to prove that ACP - BC2 = ADP - DB?.“ (Dem.) For (I. 47) AC2 = ADP + DC?

and' BCP = DB+ DC2; therefore the difference between ACP and BC2 is equal to that between ADP + DC and DB? + DC?; or AC – BC2 = ADÁ E DB - DBP.

COR. 1.-The rectangle under the sum and difference of the • two sides of a triangle, is equal to twice the rectangle under

the base and the distance of its middle point from the

perpendicular on it from the vertex. Let AB be bisected in E, then (AC + CB) (AC – CB) = 2AB ED.

(Dem.) For ACP – CBP = ADP – DB; therefore (II. 5, Cor.) AC + CB) (AC – CB) = (AD + DB) (AD – DB) = ,AB : 2ED = 2AB.ED; because AD = AE + ED = BE + ED = DB + 2ED, therefore AD – DB = 2ED, and the rectangle under AB and 2ED is equal to twice that under AB and ED. COR. 2.-If from any point in a given line a perpendicular be

drawn to it, and from any points in the latter, lines be drawn to the extremities of the former, the differences between the

squares on every two lines from the same point are equal. For these differences are all equal to the difference between the squares on the segments of the given line. COR. 3.- If. the difference between the squares on two lines

joining a point and the extremities of a given line be equal to that on other two lines similarly drawn from another point,

the line joining these two points is perpendicular to the given $14 line, if the points be on the same side of a line bisecting the

given line perpendicularly. For the segments of the given line, formed by perpendiculars upon it from these two points, must be equal (Cor. 2), or the two perpendiculars must coincide.

Schol. The first corollary to this proposition gives the most simple method of finding the segments of the side of a triangle made by a perpendicular upon it from the opposite angle, when the three sides are given; for by it ED may be found, and AD is equal to half the base, together with ED, while BD is equal to half the base diminished by ED, if the angle at B be acute, but if the angle at B be obtuse, BD is equal to

ED diminished by half the base. Hence, also, if ED be less than half the base, each of the angles at the base is acute;

but if ED be greater than half the base, one of the angles at the base is obtuse. If the sides of the triangle opposite the angles A, B, and C be called a, b, and c; the corollary gives

ED – (b + a)(b − a)

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Cor. 2.-Half the difference of two unequal lines added to

half their sum, gives the greater; and taken from half the

sum, gives the less. For AD, DB (1st fig.) are two unequal lines, AE is half their sum, and ED half their difference; and AE + ED = AD, and AE - ED = DB.

EXERCISES. 1. If from any point lines be drawn to the angular points of a rectangle, the sums of the squares on those drawn to opposite angles are equal.

2. The square on either of the sides of the right angle of a right-angled triangle is equal to the rectangle under the sum and difference of the hypotenuse and the other side.

3. The square on a perpendicular upon the hypotenuse of a right-angled triangle drawn from the opposite angle, is equal to the rectangle under the segments of the hypotenuse.

4. If a line be drawn from one of the acute angles of a rightapgled triangle to the middle of the opposite side, the square on the hypotenuse is greater than the square on that line by three times the square on half the side bisected.

5. If from the point of bisection of one of the sides of a rightangled triangle a perpendicular be drawn to the hypotenuse, the square on the third side is equal to the difference of the squares on the segments of the hypotenuse so formed.

6. If from one of the acute angles of a right-angled triangle a line be drawn to the opposite side, the squares on the hypotenuse and the segment adjacent to it, are together equal to the squares on the line so drawn, and the side on which it is drawn.

7. If a perpendicular be drawn from the right angle of a rightangled triangle on the hypotenuse, prove that the square on each of the sides is equal to the rectangle contained by the hypotenuse and its adjacent segment, and that the square on the perpendicular is equal to the rectangle contained by the segments of the hypotenuse.

8. Given two unequal lines, it is required to produce the less, so that the rectangle contained by the line thus produced and the line itself shall be equal to the square on the greater.

9. Given two straight lines, it is required to produce one of them, so that the rectangle contained by it and the produced part may be equal to the square on the other.

10. Three times the sum of the squares on the sides of a triangle, is equal to four times the sum of the squares on the three lines drawn from the angles to the middle of the opposite sides.

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