rectangle AB BC is equal to the gnomon AOH. Το each of these add XH, which is equal to the square of AC (II. 4, Cor.); therefore four times the rectangle AB BC, together with the square of AC, is equal to the gnomon AOH and the square XH. But the gnomon AOH and the square XH make up the figure AEFD, which is the square of AD; therefore four times the rectangle AB BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. COR. 1.-Hence, because AD is the sum, and AC the difference of the lines AB and BC, four times the rectangle contained by any two lines, together with the square of their difference, is equal to the square of the sum of the lines.

COR. 2. From the demonstration, it is manifest, that, since the square of CD is quadruple of the square of CB, the square of any line is quadruple of the square

of half that line.

PROPOSITION IX. THEOREM.

If a straight line be divided into two equal, and also into two unequal parts, the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.

Let the straight line AB be divided at the point C into two equal, and at D into two unequal parts; the squares of AD, DB, are together double of the squares of AC, CD.

From the point C draw CE at right angles to AB (I. 11), and make it equal to AC or CB, and join EA, EB; through D draw DF parallel to CE (I. 31), and

through F draw FG parallel to AB; and join AF. Then, because AC is equal to CE, the angle EAC is equal to the angle AEC

(I. 5); and because the angle ACE is a right angle, the two others AEC, EAC, together make one right angle (I. 32); and they are equal to one another; each of them, therefore, is half of a right angle. For the same reason, each of the angles CEB, EBC, is half a right angle; and therefore the whole AEB is a right angle. And because the angle GEF is half a right angle, and EGF a right angle,

for it is equal to the interior and opposite angle ECB (I. 29), the remaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the side EG equal to the side GF (I. 6). Again, because the angle at Bis half a right angle, and FDB a right angle, for it is equal to the interior and opposite angle ECB, the remaining angle BFD is half a right angle; therefore the angle at B is equal to the angle BFD, and the side DF to the side DB. Now, because AC is equal to CE, the square of AC is equal to the square of CE; therefore the squares of AC, CE, are double of the square of AC. But the square of EA is equal to the squares of AC, CE, because ACE is a right angle (I. 47); therefore the square of EA is double of the square of AC. Again, because EG is equal to GF, the square of EG is equal to the square of GF; therefore the squares of EG, GF, are double of the square of GF; but the square of EF is equal to the squares of EG, GF; therefore the square of EF is double of the square GF; and GF is equal to CD (I. 34); therefore the square of EF is double of the square of CD. But the square of AE is likewise double of the square of AC; therefore the squares of AE, EF, are double of the squares of AC, CD. And the square of AF is equal to the squares of AE, EF (I. 47), because AEF is a right angle; therefore the square of AF is double of the squares of AC, CD. But the squares of AD, DF, are equal to the square of AF, because the angle ADF is a right angle; therefore the squares of AD, DF, are double of the squares of AC, CD. And DF is equal to DB; therefore the squares of AD, DB, are double of the squares of AC, CD.

If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D; the squares of AD, DB, are double of the squares of AC, CD.

F

From the point C draw CE at right angles to AB (I. 11); and make it equal to AC or CB, and join AE, EB; through E draw EF parallel to AB (I. 31), and through D draw DF parallel to CE. And because the straight line EF meets the parallels EC, FD, the angles CEF, A EFD, are equal to two right angles

G

(I. 29); and therefore the angles BEF, EFD, are less than two right angles. But straight lines, which with another straight line make the interior angles upon the same side less than two right angles, do meet if produced far enough (I. 29, Cor.); therefore EB, FD, shall meet, if produced towards B, D; let them meet in G, and join AG. Then, because AC is equal to CE, the angle CEA is equal to the angle EAC (1.5); and the angle ACE is a right angle; therefore each of the angles CEA, EAC, is half a right angle (I. 32). For the same reason, each of the angles CEB, EBC, is half a right angle; therefore AEB is a right angle. And because EBC is half a right angle, DBG is also half a right angle (I. 15), for they are vertically opposite; but BDG is a right angle, because it is equal to the alternate angle DCE (I. 29); therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side DB is equal to the side DG (I. 6). Again, because EGF is half a right angle, and that the angle at F is a right angle, because it is equal to the opposite angle ECD (I. 34), the remaining angle FEG is half a right angle, and equal to the angle EGF; wherefore also the side GF is equal to the side FE. And because EC is equal to CA, the square of EC is equal to the square of CA; therefore the squares of EC, CA, are double of the square of CA. But the square of EA is equal to the squares of EC, CA (I. 47); therefore the square of EA is double of the square of AC. Again, because GF is equal to FE, the square of GF is equal to the square of FÉ; and therefore the squares of GF, FE, are double of the square of EF. But the square of EG is equal to the squares of GF, FE (I. 47); therefore the square of EG is double of the square of EF; and EF is equal to CD; wherefore the square of EG is double of the square of

CD; but it was demonstrated, that the square of EA is double of the square of AC; therefore the squares of AE, EG, are double of the squares of AC, CD. And the square of AG is equal to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD. But the squares of AD, DG, are equal to the square of AG; therefore the squares of AD, DG, are double of the squares of AC, CD. But DG is equal to DB; therefore the squares of AD, DB, are double of the squares of AC, CD.

Schol. The last two propositions are equivalent to the following:-If a straight line be bisected, and be also cut unequally, either internally or externally, the sum of the squares of the two unequal segments is equal to twice the squares of half the line and of the mean distance.

PROPOSITION XI. PROBLEM.

To divide a given straight line in medial section.

Let AB be the given straight line; it is required to divide it in medial section.

Upon AB describe the square ABDC (I. 46); bisect AC in E (I. 10), and join BE; produce CA to F, and make EF equal to EB (I. 3), and upon AF describe the square FGHA; AB is divided in H, so that the rectangle AB BH is equal to the square of AH.

H

Produce GH to K; because the straight line AC is bisected in E, and produced to the point F, the rectangle CF, FA, together with the square of AE, is equal F to the square of EF (II. 6). But EF is equal to EB; therefore the rectangle CF FA, together with the square of AE, is equal to a the square of EB; and the squares of BA, AE, are equal to the square of EB (I. 47), E because the angle EAB is a right angle; therefore the rectangle CF FA, together with the square of AE, is equal to the c squares of BA, AE; take away the square of AE, which is common to both; therefore the remaining rectangle CF FA is equal to the square of AB. Now, the figure FK is the rectangle contained by CF, FA, for AF is equal to FG, and AD is the square of AB; therefore FK is equal to

F

K

D

AD. Take away the common part AK, and the remainder FH is equal to the remainder HD. And HD is the rectangle contained by AB, BH, for AB is equal to BD, and FH is the square of AH; therefore the rectangle AB BH is equal to the square of AH. Wherefore the straight line AB is divided in H so, that the rectangle AB BH is equal to the square of AH.

PROPOSITION XII. THEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle.

Let ABC be an obtuse-angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn perpendicular to BC produced (I. 12); the square of AB is greater than the squares of AC, CB, by twice the rectangle BC CD.

Because the straight line BD is divided into two parts in the point C, the square of BD is equal to the squares of BC, CD, and twice the rectangle BC CD (II. 4); to each of these equals add the square of DA; and the squares of DB, DA, are equal to the squares of BC, CD, DA, and twice the rectangle BC CD. But the square of BA is equal to the squares of BD, DA (I. 47), because the angle at D is a right angle; and the square of CA is B equal to the squares of CD, DA. Therefore the BA is equal to the squares of BC, CA, and twice the rectangle BC CD; that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BC.CD.

square

D

of

In every triangle, the square of the side subtending any of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either